- #1

- 92

- 0

**1. A series circuit initially consists of a cell of unknown voltage, a resistor R**

a) 1.0 A

b) 2.0 A

c) 5.0 A

d) 7.5 A

_{2}of 4.0 Ω, and another resistor R_{2}of 6.0 Ω. A student measures the current through R_{2}to be 3.0 A. When a wrench that has a small resistance is dropped on the circuit, the current through R_{2}is reduced to 1.0 A. What is the current flowing through the wrench? (Assume the supply voltage remains constant.)a) 1.0 A

b) 2.0 A

c) 5.0 A

d) 7.5 A

To summarize:

R

_{1}=4.0 Ω

R

_{2}= 6.0Ω

I

_{2}initially= 3.0 A

I

_{2}with wrench= 1.0 A

Find I (current) going through the wrench.

## Homework Equations

V=IR

## The Attempt at a Solution

My answer is a) 1.0 A, but it is wrong. The correct answer is c) 5.0 A but I don't understand why.

I was told that in a series circuit, the current should be the same throughout. So in the circuit (with wrench), if the current is measured after a resistor to have a current of 1.0 A, shouldn't the current through the wrench be the same?