Question

An electric lamp whose resistance in 40ohm and a Conductor of 10ohm resistance are connected in parallel to a 12v Battery calculate the total resistance in the circuit and the total current flowing in the curcuit​

Answer 1

Answer:

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An electric lamp of resistance 20Ω and a conductor of resistance 4Ω are 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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Solution

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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=

Total Resistance

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

Answer:

1.5 ampere

Explanation:

R1 (lamp) = 40 ohm

R2 (conductor) = 10 ohm

R (total resistance) = 1/ (1/40 + 1/10) {because of parallel connection}

= 1/ (5/40)

= 40/5 = 8 ohm

Total resistance (R) = 8 ohm

Voltage = 12 V

V=IR

I=V/R = 12/8 = 1.5 amp