The supply voltage to room is 120 V. The resistance of the lead wires is `6Omega`. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?

Options

A. 2.9 Volt
B. 13.3 Volt
C. 10.04 Volt
D. zero volt

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the resistance of the 60 W bulb
The power (P) of the bulb is given as 60 W and the supply voltage (V) is 120 V. We can use the formula for power:
\[ P = \frac{V^2}{R} \]
Rearranging this gives:
\[ R = \frac{V^2}{P} \]
Substituting the values:
\[ R_{\text{bulb}} = \frac{120^2}{60} = \frac{14400}{60} = 240 \, \Omega \]

### Step 2: Calculate the resistance of the 240 W heater
Using the same formula for the heater:
\[ P = 240 \, W \]
\[ R_{\text{heater}} = \frac{120^2}{240} = \frac{14400}{240} = 60 \, \Omega \]

### Step 3: Calculate the voltage drop across the bulb when only the bulb is on
When only the bulb is on, the total resistance in the circuit is the sum of the bulb's resistance and the lead wire's resistance:
\[ R_{\text{total}} = R_{\text{bulb}} + R_{\text{wire}} = 240 + 6 = 246 \, \Omega \]
The voltage drop across the bulb can be calculated using the voltage divider rule:
\[ V_{\text{bulb}} = V \times \frac{R_{\text{bulb}}}{R_{\text{total}}} = 120 \times \frac{240}{246} \]
Calculating this gives:
\[ V_{\text{bulb}} = 120 \times \frac{240}{246} \approx 117.07 \, V \]

### Step 4: Calculate the equivalent resistance when both the bulb and heater are on
When the bulb and heater are on, they are in parallel. The equivalent resistance \( R_{\text{eq}} \) of the bulb and heater can be calculated as:
\[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_{\text{bulb}}} + \frac{1}{R_{\text{heater}}} \]
Substituting the values:
\[ \frac{1}{R_{\text{eq}}} = \frac{1}{240} + \frac{1}{60} \]
Finding a common denominator (240):
\[ \frac{1}{R_{\text{eq}}} = \frac{1}{240} + \frac{4}{240} = \frac{5}{240} \]
Thus,
\[ R_{\text{eq}} = \frac{240}{5} = 48 \, \Omega \]

### Step 5: Calculate the new voltage drop across the bulb with both devices on
Now, the total resistance in the circuit is:
\[ R_{\text{total}} = R_{\text{eq}} + R_{\text{wire}} = 48 + 6 = 54 \, \Omega \]
Using the voltage divider rule again:
\[ V'_{\text{bulb}} = V \times \frac{R_{\text{eq}}}{R_{\text{total}}} = 120 \times \frac{48}{54} \]
Calculating this gives:
\[ V'_{\text{bulb}} = 120 \times \frac{48}{54} \approx 106.67 \, V \]

### Step 6: Calculate the decrease in voltage across the bulb
The decrease in voltage across the bulb when the heater is switched on is:
\[ \Delta V = V_{\text{bulb}} - V'_{\text{bulb}} = 117.07 - 106.67 \approx 10.4 \, V \]

### Final Answer
The decrease in voltage across the bulb when the heater is switched on is approximately **10.4 V**.

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