A coil of resistance `11Omega` and inductance `0.1H` is connected to a 110V d.c. mains. Find
(i) the current finally established in the coil.
(ii) the voltage used in overcoming the resistance when the rising current is 3A, and
(iii) the rate at which the current is rising at that instant.

Answer

To solve the problem step by step, we will address each part of the question systematically.

### Given Data:
- Resistance, \( R = 11 \, \Omega \)
- Inductance, \( L = 0.1 \, H \)
- Voltage, \( \epsilon = 110 \, V \)

### (i) Find the final current established in the coil.

The final current (\( I_0 \)) in a DC circuit with resistance and inductance can be calculated using Ohm's law:

\[ I_0 = \frac{\epsilon}{R} \]

Substituting the given values:

\[ I_0 = \frac{110 \, V}{11 \, \Omega} = 10 \, A \]

**Final Answer for (i):** The final current established in the coil is \( 10 \, A \).

### (ii) Find the voltage used in overcoming the resistance when the rising current is \( 3 \, A \).

The voltage across the resistance (\( V_R \)) can be calculated using Ohm's law:

\[ V_R = I \cdot R \]

Where \( I \) is the current at that instant. Substituting \( I = 3 \, A \):

\[ V_R = 3 \, A \cdot 11 \, \Omega = 33 \, V \]

**Final Answer for (ii):** The voltage used in overcoming the resistance when the rising current is \( 3 \, A \) is \( 33 \, V \).

### (iii) Find the rate at which the current is rising at that instant.

The equation for the current in an RL circuit is given by:

\[ I(t) = I_0 \left(1 - e^{-\frac{R t}{L}}\right) \]

We need to find the time \( t \) when the current \( I = 3 \, A \):

Setting \( I(t) = 3 \, A \):

\[ 3 = 10 \left(1 - e^{-\frac{R t}{L}}\right) \]

Rearranging gives:

\[ \frac{3}{10} = 1 - e^{-\frac{R t}{L}} \]

\[ e^{-\frac{R t}{L}} = 1 - \frac{3}{10} = \frac{7}{10} \]

Taking the natural logarithm on both sides:

\[ -\frac{R t}{L} = \ln\left(\frac{7}{10}\right) \]

Thus,

\[ \frac{R t}{L} = -\ln\left(\frac{7}{10}\right) \]

Now substituting \( R = 11 \, \Omega \) and \( L = 0.1 \, H \):

\[ t = \frac{-L \ln\left(\frac{7}{10}\right)}{R} = \frac{-0.1 \ln\left(\frac{7}{10}\right)}{11} \]

Now we need to differentiate the current equation to find the rate of change of current (\( \frac{dI}{dt} \)):

\[ \frac{dI}{dt} = \frac{I_0 R}{L} e^{-\frac{R t}{L}} \]

Substituting \( I_0 = 10 \, A \), \( R = 11 \, \Omega \), and \( L = 0.1 \, H \):

\[ \frac{dI}{dt} = \frac{10 \cdot 11}{0.1} e^{-\frac{R t}{L}} = 1100 e^{-\frac{R t}{L}} \]

From our previous calculation, we know \( e^{-\frac{R t}{L}} = \frac{7}{10} \):

\[ \frac{dI}{dt} = 1100 \cdot \frac{7}{10} = 770 \, A/s \]

**Final Answer for (iii):** The rate at which the current is rising at that instant is \( 770 \, A/s \).

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