Relation Between Power and Resistance

1.0What is Power in an Electrical Circuit?

2.0Electrical Power

Energy liberated per second in a device is called its power. The electrical power P delivered or consumed by an electrical device is given by P = Vi, where V = Potential difference across the device and i = Current.

Power (P) = V·dq/dt = Vi

If the current enters the higher potential point of the device then electric power is consumed by it (i.e. acts as load). If the current enters the lower potential point then the device supplies power (i.e. acts as source).

• Unit of Power: Watt (W).
• 1 Watt = 1 Joule of energy consumed per second.

3.0Heating Effect of Current

Cause of Heating: The potential difference applied across the two ends of conductor sets up electric field. Under the effect of electric field, electrons accelerate and as they move, they collide against the ions and atoms in the conductor, the energy of e⁻ transferred to the atoms and ions appears as heat.

4.0Relation between Power and Resistance

Joule’s Law of Heating

Let voltage ( V ) be applied across the conductor of resistance ( R ) and charge ( dq ) flows through it in time ( dt ), then work done by the battery is

dw = dqV

$=i,dt\cdot V$

$dw=iVdt=i^{2}Rdt$

Since heat energy released from the conductor is equal to work done by the battery unit,

$H=\text{Heat released}=\int i^{2}R,dt$

If ( i ) = constant,

$H=i^{2}Rt=\frac{V^2}{R}t=VIt$

Heat produced in a conductor does not depend upon the direction of current.

SI UNIT: joule

1 kWh = ( $3.6\times 10^6$ ) joule

1 BTU (British Thermal Unit) = 1055 J

Power delivered to the load resistance, P = $\frac{dW}{dt}$

$P=i^{2}R=\frac{V^2}{R}=VI$

5.0Fusing of an Electrical Device

Every electrical device like bulb, heater, geyser, etc. comes with a design/rated voltage ( $V_0$ ) and rated power ( $P_0$ ).

If applied voltage ( V ) across a device becomes greater than ( $V_0$ ), then it will not work or we can say the device will be fused.

Resistance of an electrical device can be found out using

$P_0=\frac{V_0^2}{R}$

i.e.

$R=\frac{V_0^2}{P_0}$

If ( V ) is the applied voltage across the device and ( P ) is the power consumed by it, then $P=\frac{V^2}{R}={\left(\frac{V}{V_0}\right)}^2{P}_0$

6.0Home Appliances (Bulb, Heater, Geyser)

Equivalent power in combination of Bulbs

(i) Series combination

$Let2bulbswit{V}_{0}but{P}_{0_1}>P_{0_2}areconnectedinseries.$

$Here,P_{0_1}>P_{0_2}\Rightarrow R_1<R_2$

$Now,sinceseries\Rightarrow P_1=I^2{R}_1\text{and}{P}_2=I^2{R}_2$

$\therefore P_1<P_2$

$Inseriescombinationofbulbs,sinceBrightness\propto Powerconsumedbybulb\propto R\propto \frac{1}{P_{\text{rated}}},bulboflesserwattagewillshinemore.$

$P=\text{total power of combination}=\frac{V^2}{R_1+R_2}=\frac{V^2}{\frac{V_0^2}{P_{0_1}}+\frac{V_0^2}{P_{0_2}}}={\left(\frac{V}{V_0}\right)}^2\left(\frac{P_{0_1}{P}_{0_2}}{P_{0_1}+P_{0_2}}\right)$

$IfV=V_0,then\frac{1}{P}=\frac{1}{P_{0_1}}+\frac{1}{P_{0_2}}$

(ii) Parallel combination

Similarly, let two bulbs with $V_{0}but{P}_{0_1}>P_{0_2}$ are connected in parallel.

$Since,P_{0_1}>P_{0_2}$

$\therefore R_1<R_2$

$Now{P}_1=\frac{V^2}{R_1};P_2=\frac{V^2}{R_2}$

$\therefore P_1>P_2$

In parallel combination of bulbs since Brightness $\propto$ Power consumed by bulb $\propto I\propto \frac{1}{R}\propto P_{\text{rated}}$, so bulb of greater wattage will shine more.

$P_{\text{total}}=\frac{V^2}{R_1}+\frac{V^2}{R_2}={\left(\frac{V}{V_0}\right)}^2[P_1+P_2]$

$IfV=V_0$

$P_{\text{total}}=P_1+P_2$

7.0Single Loop Circuit and Power Output, R for Maximum Power Output and Maximum Current

Electric current in resistance

In a resistor current flows from high potential to low potential

• High Potential is represented by positive sign (+)
• Low potential is represented by negative sign (-)

$V_A-V_B=iR$

If V1 > V2
Then current will flow from A to B, and i = (V1 − V2) / R

If V1 < V2
Then current will go from B to A, and i = (V2 − V1) / R

Maximum Power Output

In the given electric circuit.

If we assume that potential at A is zero then potential at B is $\epsilon -Ir$

Now since the connecting wires are of zero resistance

$\therefore V_D=V_A=0\Rightarrow V_C=V_B=\epsilon -Ir$

Now current through CD is also I (because it's in series with the cell)

$\therefore I=\frac{V_C-V_D}{R}=\frac{(\epsilon -Ir)-0}{R}\Rightarrow I=\frac{\epsilon }{r+R}$

Power output in R

$\Rightarrow P=I^{2}R=\frac{{\epsilon }^2}{(r+R{)}^2}R$

$\text{For maximum power supply}\Rightarrow \frac{dP}{dR}=\frac{{\epsilon }^2}{(r+R{)}^2}-\frac{2{\epsilon }^{2}R}{(r+R{)}^3}=\frac{{\epsilon }^2}{(R+r{)}^3}[R+r-2R]$

$\text{For maximum power supply}=\frac{dP}{dR}=0\Rightarrow r+R-2R=0\Rightarrow r=R$

Here for maximum power output external resistance should be equal to internal resistance.

Now, graph between 'P' and ‘R’.

Maximum power output at R

$P_{\max }=\frac{{\epsilon }^2}{4R}\Rightarrow I=\frac{\epsilon }{r+R}=\frac{\epsilon }{2R}$

Now, from the graph we can see that for a given power output there exists two values of external resistance.

$P=\frac{{\epsilon }^{2}R}{(r+R{)}^2}$

$P(r^2+2rR+R^2)={\epsilon }^{2}R$

$R^2+\left(2r-\frac{{\epsilon }^2}{P}\right)R+r^2=0$

Above quadratic equation in R has two roots R1 and R2 for given values $\epsilon$, P and r such that

$$\begin{gathered} R_1{R}_2=r^2(productofroots) \\ {r}^2=R_1{R}_2 \end{gathered}$$

8.0Electrical Resistance

The property of a substance by virtue of which it opposes the flow of electric current through it is termed as electrical resistance. Electrical resistance depends on the size, geometry, temperature and internal structure of the conductor.

We have,

$I=\frac{nA{e}^2\tau }{m\ell }V$

$I=\frac{V}{R}$

$R=\frac{m\ell }{nA{e}^2\tau }$

Hence,

$R=\frac{m}{n{e}^2\tau }\frac{\ell }{A}$

So, here

$R=\frac{\rho \ell }{A}\Rightarrow V=I\times \frac{\rho \ell }{A}$

$\Rightarrow \frac{V}{\ell }=\frac{I}{A}\rho \Rightarrow E=J\rho \Rightarrow J=\frac{I}{A}=\text{current density}$

$\rho$ is called resistivity (it is also called specific resistance), and $\rho =\frac{m}{n{e}^2\tau }=\frac{1}{\sigma }$

$\sigma$ is called conductivity. S.I. unit of resistivity is ohm-m $(\Omega \text{-}m)$.

S.I. unit of conductivity is ${\Omega }^{-1}\text{-}{m}^{-1}$ also called siemens.

Dependence of Resistance on Various Factors

From previous discussion we have:
R = ρ l / A = (m / (n e² τ)) × (l / A)

Therefore, resistance depends on

(1) length of the conductor (R ∝ l)
(2) area of cross-section of the conductor (R ∝ 1/A)
(3) nature of material of the conductor R = ρ l / A

Results:

(a) On stretching a wire (volume constant), if length of wire is taken into account then $\frac{R_1}{R_2}=\frac{{\ell }_1^2}{{\ell }_2^2}$

(b) If radius of cross-section is taken into account then $\frac{R_1}{R_2}=\frac{r_2^4}{r_1^4}$, where, $R_1$ and $R_2$ are initial and final resistances and $l_1,l_2$ are initial and final lengths and $r_1,r_2$ are initial and final radii respectively.

(c) Effect of percentage change in length of wire$\frac{R_2}{R_1}=\frac{{\ell }^2{\left[1+\frac{x}{100}\right]}^2}{{\ell }^2},\text{where }\ell \text{ - original length and }x\text{ - \% increment}$

If x is quite small (say < 5%) then % change in R is $\frac{R_2-R_1}{R_1}\times 100=\left(\frac{{\left(1+\frac{x}{100}\right)}^2-1}{1}\right)\times 100\cong 2x\%$

Resistance is the property of a material that opposes the flow of electric current. It depends on the material, length, and cross-sectional area of the conductor.

• Unit of Resistance: Ohm (Ω).
• High resistance → Less current flows.
• Low resistance → More current flows.

Mathematically, resistance is given by: $R=\frac{V}{I}$

Role of Resistance in Power Dissipation

When current flows through a resistor, electrical energy is converted into heat energy due to collisions of electrons with atoms of the conductor. This is called the Joule heating effect.

The heat generated per second is the power dissipated by the resistor: $P=I^{2}R$

Thus, resistance plays a direct role in how much power is consumed or lost in a circuit.

9.0Relation between Power and Resistance Formula

In Physics, power is defined as the rate at which electrical energy is consumed or converted into other forms of energy, such as heat, light, or mechanical work. For an electric circuit, power depends on three key quantities: voltage (V), current (I), and resistance (R).

The general formula for electrical power is:

P=VI

Where:

• P = Power (in watts, W)
• V = Voltage across the conductor (in volts, V)
• I = Current flowing through the conductor (in amperes, A)

Now, using Ohm’s Law:

V=IR

we can derive the relation between power and resistance.

By substituting V=IR into P=VI: $P=I^{2}R$

Alternatively, substituting $I=\frac{V}{R}:P=\frac{V^2}{R}$

So, the formulas relating power and resistance are:

• $P=I^{2}R$
• $P=\frac{V^2}{R}$

These two equations are the fundamental expressions connecting power and resistance in an electric circuit.

10.0Power and Resistance Relation

The relation between power and resistance depends on whether voltage or current is constant in the circuit. Let’s explore both cases.

Case I: Power in terms of Current (Constant Current Supply)

If current remains constant, the relation is: $P=I^{2}R$

Here, power is directly proportional to resistance.

• If resistance increases, power increases.
• If resistance decreases, power decreases.

Example: A current of 2 A flows through resistors.

• For R=5 Ω: $P=I^{2}R=(2{)}^2\times 5=20W$
• For R=10 Ω: $P=I^{2}R=(2{)}^2\times 10=40W$

Clearly, higher resistance results in higher power dissipation under constant current.

Case II: Power in terms of Voltage (Constant Voltage Supply)

If voltage remains constant, the relation is:

$P=\frac{V^2}{R}$

Here, power is inversely proportional to resistance.

• If resistance increases, power decreases.
• If resistance decreases, power increases.

Example: A voltage of 20 V is applied across resistors.

• For R=5 Ω: $P=\frac{V^2}{R}=\frac{400}{5}=80W$
• For R=10 Ω: $P=\frac{V^2}{R}=\frac{400}{10}=40W$

Thus, higher resistance reduces the power dissipation under constant voltage.

Practical Significance of the Relation

• Electrical Appliances: Devices like heaters and bulbs work on the principle of , where resistance wires convert electrical energy into heat/light.
• Power Transmission: In long-distance transmission lines, minimizing resistance is crucial to reduce unwanted power loss.
• Circuit Design: Engineers choose suitable resistors to control power usage and prevent overheating.

11.0Solved Examples

Question 1: The dimensions of a conductor of specific resistance ρ are shown below. Find the resistance of the conductor across AB, CD and EF.

Solution:

For a condition
R = ρl / A = (Resistivity × length) / (Area of cross section)

$$\begin{gathered} R_{AB}=\rho c/ab \\ {R}_{CD}=\rho b/ac \\ {R}_{EF}=\rho a/bc \end{gathered}$$

Question 2: If a wire is stretched to double its length, find the new resistance if original resistance of the wire was R.

Solution: As we know that R = ρl / A ⇒ in case R′ = ρl′ / A′

l′ = 2l

A′l′ = Al (Volume of the wire remains constant)

A′ = A / 2

⇒ R′ = (ρ × 2l) / (A / 2) = 4ρl / A = 4R

Question 3: The wire is stretched to increase the length by 1%. Find the percentage change in the resistance.

Solution: As we know that,
R = ρl / A

⇒ ΔR / R = Δρ / ρ + Δl / l − ΔA / A and Δl / l = − ΔA / A

ΔR / R = 0 + 1 + 1 = 2

Hence percentage increase in the resistance = 2%.

Note: Above method is applicable when % change is very small.

Question 4: If bulb rating is 100 watt and 220 V then determine:
(a) Resistance of filament
(b) Current through filament
(c) If bulb operate at 110-volt power supply then find power consumed by bulb.

Solution: Bulb rating is 100 W and 220 V bulb means when 220 V potential difference is applied between the two ends then the power consumed is 100 W

Here, V = 220 Volt
P = 100 W

V² / R = 100; So, R = 484 Ω

Since resistance depends only on material hence it is constant for bulb

I = V / R = 220 / 484 = 5/11 Amp

Power consumed at 110 V

Power consumed = (110 × 110) / 484 = 25 W

Question 5: Two wires of same mass, having ratio of lengths 1 : 2, density 1 : 3, and resistivity 2 : 1, are connected one by one to the same voltage supply. The rate of heat dissipation in the first wire is found to be 10 W. Find the rate of heat dissipation in the second wire.

Solution:

Given,

$\frac{l_1}{l_2}=\frac{1}{2},\frac{d_1}{d_2}=\frac{1}{3},\frac{{\rho }_1}{{\rho }_2}=\frac{2}{1}$

$$\begin{gathered} M=A_1{l}_1{d}_1=A_2{l}_2{d}_2 \\ \therefore \frac{P_2}{P_1}=\frac{V^2/R_2}{V^2/R_1}=\frac{R_1}{R_2} \\ =\frac{{\rho }_1{l}_1{A}_2}{{\rho }_2{l}_2{A}_1}=\frac{{\rho }_1{l}_1^2{d}_1}{{\rho }_2{l}_2^2{d}_2} \\ =\frac{2}{1}\times {\left(\frac{1}{2}\right)}^2\times \frac{1}{3}=\frac{1}{6}or{P}_2=\frac{P_1}{6}=\frac{10}{6}=\frac{5}{3}\text{W} \end{gathered}$$

Question 6: In the figure shown B₁, B₂ and B₃ are three bulbs rated as (200 V, 50 W), (200 V, 100 W) and (200 V, 25 W) respectively. Find the current through each bulb and which bulb will give more light?

Solution:

$R_1=\frac{(200{)}^2}{50};R_2=\frac{(200{)}^2}{100};R_3=\frac{(200{)}^2}{25}$

The current flowing through each bulb is

$=\frac{200}{R_1+R_2+R_3}=\frac{200}{(200{)}^2\left[\frac{2+1+4}{100}\right]}=\frac{100}{200\times 7}=\frac{1}{14}A$

Since, $R_3>R_1>R_2$

• Power consumed by bulb = i²R
• If the resistance is of higher value then it will give more light.
• Here Bulb $B_3$ will give more light.

Question 7: If two heater coils boil same amount of water in time t₁ and t₂ separately, find time taken by the combination of these coils if (i) they are connected in series and (ii) they are connected in parallel?

Solution:

(i) If connected in series then total power P to water will be given as

1 / P = 1 / P₁ + 1 / P₂

tₛ / H = t₁ / H + t₂ / H

tₛ = t₁ + t₂

(ii) Similarly, in parallel, we know total power, P = P₁ + P₂

So, H / tₚ = H / t₁ + H / t₂

tₚ = (t₁ t₂) / (t₁ + t₂)

Question 8: A heater coil is rated 100 W, 200 V. It is cut into two identical parts. Both parts are connected together in parallel, to the same source of 200 V. Calculate the energy liberated per second in the new combination.

Solution:

P = V² / R

R = V² / P = (200)² / 100 = 400 Ω

Resistance of half piece = 400 / 2 = 200 Ω

Resistance of pieces connected in parallel = 200 / 2 = 100 Ω

Energy liberated per second = P = V² / R = (200 × 200) / 100 = 400 W