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Fundamentals Of Physics
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Short Answer

The current in a single-loop circuit with one resistance R is 5.0 A. When an additional resistance of 2.0 Ω is inserted in series with R, the current drops to 4.0 A. What is R?

The value of R is.

See the step by step solution

Step by Step Solution

Step 1: Given

I=5.0 AR'=2.0 ΩI'=4.0 A

Step 2: Determining the concept

Write for emf of the battery when the circuit has only resistance R and when R’ is inserted in the circuit using Ohm’s law. Equating them will give the value of R.

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

Formulae are as follow:

I=VR

Where, I is current, V is voltage, R is resistance.

Step 3: Determining the value of R

Let the battery be emf V. Then, Ohm’s law gives,

V=IRV=5R

When R’ is inserted in the circuit in series with R, the total resistance of the circuit becomes R+R’.

Then emf is given by,

V=I'R+R'V=4R+2

Equating the equations 1) and 2) gives,

5R=4R+8R=8.0 Ω

Hence, the value of R is 8.0 Ω.

Therefore, using Ohm’s law, the unknown resistance in the circuit can be found.

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