Question

. An electric lamp whose resistance is 8 ohms and a conductor of 2 ohms resistance are connected in series to a 5 volt battery. Calculate (A) The total resistance of the circuit. (B) The current through the circuit. (C) The potential difference across the electric lamp and the conductor. (1+1+1)​

Answer 1

Given : An electric lamp whose resistance is 8 ohms and a conductor of 2 ohms resistance are connected in series to a 5 volt battery .

To Find : Calculate

a) the total resistance of the circuit

b) the current through the circuit

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

Solution:

Resistance in series gets added

hence total resistance of the circuit   = 8 + 2 = 10 Ω

I  = V/R

V = 5 V

R = 10  Ω

=> I = 5/10

=> I = 0.5 A

potential difference  = IR

potential difference across the electric lamp  =  0.5 * 8  = 4V

potential difference across the  conductor​   =  0.5 * 2  = 1 V

a) the total resistance of the circuit  = 10  Ω

b) the current through the circuit  = 0.5 A

c) the potential difference across the electric lamp = 4 V and the conductor​ = 1 V

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Answer 2

Answer:

Total resistance of circuit

=8ohm+2ohm=10ohm

again,

V=5V

R=10ohm

Then,

I=V/R

I=5/10

I=0.5A

Now,

V=IR

V=0.5A*8ohm

V=4V

and,

V=IR

V=0.5A*2ohm

V=1V

Hence,

total resistance of circuit is 10ohm.

total current through the circuit is 0.5A.

potential difference across the electric lamp and the conductor is 4V and 1V.

I think you understand my answer