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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.

Respuesta :

Answer:

Value of [tex]R_1=30.4ohm[/tex]

Explanation:

We have given

In first case resistance is [tex]R_1[/tex] and current is 1.8 A

Let the potential difference is v

So [tex]1.8=\frac{v}{R_1}[/tex]----eqn 1

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

So [tex]1.6=\frac{v}{R+3.8}[/tex]----eqn 2

From eqn 1 and eqn 2

[tex]1.8R_1=1.6R_1+6.08[/tex]

[tex]R_1=30.4ohm[/tex]

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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