Question

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w. An electric lamp whose resistance is 2012 and a resistor of 49 resistance are
connected in parallel in an electric circuit. (A battery of 6V is connected in the
circuit.)
( a ) Calculate the total resistance of the circuit.
(b) Calculate the electric current passing through the circuit.
1) Calculate the potential difference across the electric lamp.​

Answer 1

Answer:

An electric lamp of resistance 20Ω and a conductor of resistance 4Ω 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.

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Answer

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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.