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Ask one question at a time. As per the prevailing rules I am to answer only one question at a time.
E.M.F. of the circuit = 9 Volts.
In order to find the total effective resistance of the circuit, the internal resistance of the light bulb has to be worked out first.
For the light bulb `L_1` , ∆V = 6V, I = 0.5A, therefore, R = ∆V/I = 6/0.5 = 12 ohms.
Total effective resistance of the circuit = (`R_1` +12) ohms (ignoring the internal resistance of the 9V cell).
Therefore, current passing through the light bulb `L_1` (and also through the resistor `R_1` )
= `9/(R_1+12)` Amperes.
I'm Sorry! The power of the light bulb L_1 is given, which is 0.5 W.
Watt = Volt × Current
0.5 = 6 × i
`rArr` i = 0.5/6 = 0.083 Amperes(approx).
Therefore, the current through the bulb is 0.083 amperes.
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