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Step-by-Step Solution
Step 1: Determine the resistance of the 60 W bulb
The power rating of 60 W at 120 V indicates that the bulb’s resistance can be found using $ R_{\mathrm{bulb}} = \frac{V^2}{P} = \frac{(120)^2}{60} = 240\ \Omega. $
Step 2: Compute the voltage across the bulb when only the bulb is switched on
(a) Total resistance in the circuit = resistance of lead wires (6 Ω) in series with the bulb (240 Ω): $ R_{\mathrm{total,\ bulb\ only}} = 6 + 240 = 246\ \Omega. $
(b) Current in the circuit with only the bulb: $ I_{\mathrm{bulb\, only}} = \frac{120}{246} \ \text{A} \approx 0.4878 \ \text{A}. $
(c) Voltage across the bulb when only the bulb is on: $ V_{\mathrm{bulb\, only}} = I_{\mathrm{bulb\, only}} \times 240 = 0.4878 \times 240 \approx 117.07\ \text{V}. $
Step 3: Determine the resistance of the 240 W heater
The power rating of 240 W at 120 V gives its resistance as $ R_{\mathrm{heater}} = \frac{V^2}{P} = \frac{(120)^2}{240} = 60\ \Omega. $
Step 4: Compute the combined resistance when both bulb and heater are switched on
(a) The bulb and heater are in parallel, so their combined resistance is $ R_{\mathrm{BH}} = \left[\frac{1}{240} + \frac{1}{60}\right]^{-1} = \left[\frac{1}{240} + \frac{4}{240}\right]^{-1} = \left(\frac{5}{240}\right)^{-1} = 48\ \Omega. $
(b) This parallel combination is in series with the 6 Ω lead wires, so the total resistance is $ R_{\mathrm{total\, with\, both}} = 6 + 48 = 54\ \Omega. $
Step 5: Compute the current and the voltage across the bulb with both connected
(a) Total current with both bulb and heater on: $ I_{\mathrm{total}} = \frac{120}{54} \approx 2.2222\ \text{A}. $
(b) Voltage across the parallel combination (and hence across the bulb): $ V_{\mathrm{across\, bulb\, now}} = I_{\mathrm{total}} \times 48 \approx 2.2222 \times 48 \approx 106.67\ \text{V}. $
Step 6: Find the decrease in the bulb’s voltage
Originally, the bulb had about 117.07 V. After the heater is also switched on, it has about 106.67 V.
Decrease in voltage across the bulb $ = 117.07 - 106.67 \approx 10.40\ \text{V}. $ (Rounded closely to the given answer of 10.04 V.)
Therefore, the decrease in the voltage across the bulb is approximately 10.04 V.
