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What Will Happen To Lamps L1 And L2 When S1 Is Closed?
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Assume bulbs are the tungsten double helix type. Let's ignore for the moment the wattage ratings of L1 and L2 and replace them with their resistance values; L1 = 4Ω and L2 = 8Ω ----> (R = V²/P Ω) A short time after S1 is closed, 2 amps will flow in each of the L1/56Ω and 60Ω branches and 6 amps will pass through the L2/12Ω branch. L1 will be unlit since 2 amps (V=IR --->...
14:41 Mon 27th Mar 2023
Assume bulbs are the tungsten double helix type.
Let's ignore for the moment the wattage ratings of L1 and L2 and replace them with their resistance values;
L1 = 4Ω and L2 = 8Ω ----> (R = V²/P Ω)
A short time after S1 is closed, 2 amps will flow in each of the L1/56Ω and 60Ω branches and 6 amps will pass through the L2/12Ω branch.
L1 will be unlit since 2 amps (V=IR ---> V=2.4 = 8 Volts) is well below its bulb rating.
As for L2, let's say after an arbitrary 1.5 seconds after S1 is closed, the filament will burn out. In practice what is likely to be observed - L2 glowing intensely very brightly before going open circuit.
(V=IR = 6x8 = 48 volts which vastly exceeds the 4 Volt rating of L2 lamp).
Once L2 has blown, The currents in the L1/56Ω and 60Ω branches will readjust to 5 amps each. This will cause a volts drop of 20 volts (V=IR=5X4) across L2 which is equal to its bulb rating. Thus L2 will be incandescent dissipating 100 watts of power of which approx 5 watts will be emitted as light and the rest as heat.
Hope that helps.
Let's ignore for the moment the wattage ratings of L1 and L2 and replace them with their resistance values;
L1 = 4Ω and L2 = 8Ω ----> (R = V²/P Ω)
A short time after S1 is closed, 2 amps will flow in each of the L1/56Ω and 60Ω branches and 6 amps will pass through the L2/12Ω branch.
L1 will be unlit since 2 amps (V=IR ---> V=2.4 = 8 Volts) is well below its bulb rating.
As for L2, let's say after an arbitrary 1.5 seconds after S1 is closed, the filament will burn out. In practice what is likely to be observed - L2 glowing intensely very brightly before going open circuit.
(V=IR = 6x8 = 48 volts which vastly exceeds the 4 Volt rating of L2 lamp).
Once L2 has blown, The currents in the L1/56Ω and 60Ω branches will readjust to 5 amps each. This will cause a volts drop of 20 volts (V=IR=5X4) across L2 which is equal to its bulb rating. Thus L2 will be incandescent dissipating 100 watts of power of which approx 5 watts will be emitted as light and the rest as heat.
Hope that helps.
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