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lesya [120]
3 years ago
8

A loop circuit has a resistance of R1 and a current of 1.8 A. The current is reduced to 1.6 A when an additional 3.8 Ω resistor

is added in series with R1. What is the value of R1? Assume the internal resistance of the source of emf is zero. Answer in units of Ω.
Physics
2 answers:
Allisa [31]3 years ago
6 0

Answer:

Value of R_1=30.4ohm

Explanation:

We have given

In first case resistance is R_1 and current is 1.8 A

Let the potential difference is v

So 1.8=\frac{v}{R_1}----eqn 1

In second case resistance is R_1+3.8 and current is 1.6 A and potential difference will be as it is a series connection

So 1.6=\frac{v}{R+3.8}----eqn 2

From eqn 1 and eqn 2

1.8R_1=1.6R_1+6.08

R_1=30.4ohm

alisha [4.7K]3 years ago
6 0

Answer:

30.4 ohm

Explanation:

Let V be the potential difference.

When R1 is in the circuit.

V = 1.8 x R1 ... (1)

Now new resistance is R = R1 + 3.8

V = 1.6 x (R1 + 3.8) .... (2)

By comparing (1) and (2), we get

1.8 R1 = 1.6 (R1 + 3.8)

1.8 R1 - 1.6 R1 = 6.08

R1 = 30.4 ohm

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