Question

1. In the circuit shown, R - 10.on, Rac 2000 R and R₂ = . 3 2. The current through he is I, 5.004. The battery supplies an un
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Answer #1

Given is:-

R1 = 102, R2 = 202, R3 = 13.332 11 = 5A

Now,

two resistors which are connected in parallel, their equivalnet resistance is given by
R1 x R2 R12 = R1 + R2

thus

10 x 20 R12 = 10 + 20 100 = 6.6732

now the equivalent resistance of the ckt is

Req = R12 + R3

Reg= 6.67 + 13.33

which gives us

Req = 2012

Part - b

Current divides in the inverse raiton of resistances therefore the current through R2 will be

10 20 * 54

which gives us

12 = 2.54

Part - c

the total current in the ckt is

I = 5 +2.5 = 7.5A

thus the emf of the battery is
E=1 x Reg

or

E = 7.5A x 200

which gives us

E = 150V

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