Question

An electric lamp of resistance 40Ω and a conductor of resistance 8Ω re connected to
a 6V battery as shown in the circuit Calculate:
(a) The total resistance of the circuit
(b) The current through the circuit
(c)The potential difference across the (i) Electric lamp and 
(ii) Conductor, and 
(d) Power of the lamp.

Answer 1

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• a). the total resistance of the circuit =20Ω+4Ω=24Ω

• b). The current through the circuit= current through the bulb= current through the conductor.

The current through the circuit= the voltage of the batteryvoltage applied.

the current through the circuit =246amp

The current through the circuit = 0.25amp

• c). The potential difference across the lamp and the conductor.

(i) The potential difference across the electrical lamp =0.25×20=5volt

(ii) The potential difference conductor =0.25×4=1 volt

• d).  Power can be calculated as current through the lampvoltage across the lamp

     power=0.255→20W.

Answer 2

Answer:

\huge \underline \green{\sf{ Answer :- }}

Answer:−

a). the total resistance of the circuit =20Ω+4Ω=24Ω

b). The current through the circuit= current through the bulb= current through the conductor.

The current through the circuit= the voltage of the batteryvoltage applied.

the current through the circuit =246amp

The current through the circuit = 0.25amp

c). The potential difference across the lamp and the conductor.

(i) The potential difference across the electrical lamp =0.25×20=5volt

(ii) The potential difference conductor =0.25×4=1 volt

d). Power can be calculated as current through the lampvoltage across the lamp

power=0.255→20W.

Explanation:

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