Question

1) An electric lamp of resistance 20 Ω and a conductor of resistance 4 Ω are 5m connected to a 6 V 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 (ii) conductor (d) power of the lamp






Answer please

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Answer 1

Explanation:

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 battery

voltage applied

.

the current through the circuit =

24

6

amp

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 lamp

voltage across the lamp

power=

0.25

5

→20W.

Answer 2

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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 battery

voltage applied

.

the current through the circuit =

24

6

amp

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 lamp

voltage across the lamp

power=

0.25

5

→20W.