Question

An electric lamp, whose resistance is 26 Ω, and a conductor of 4 Ω

resistance are connected to a 6 V battery (Fig. 12.9). Calculate (a) the

total resistance of the circuit, (b) the current through the circuit, and

(c) the potential difference across the electric lamp and conductor.​

Answer 1

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.

Explanation:

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Answer 2

Explanation:

total resistance will be 30 ohm and current through each will be same if both are taken in series and will be I = 0.2 A also current through circuit will be 0.2A

total potential diff. across lamp will be 5.2 V and through conductor 0.8 V.