   Chapter 18, Problem 14P

Chapter
Section
Textbook Problem

A battery with ε = 6.00 V and no internal resistance supplies current to the circuit shown in Figure P18.14. When the double-throw switch S is open as shown in the figure, the current in the battery is 1.00 mA. When the switch is closed in position a, the current in the battery is 1.20 mA. When the switch is closed in position b, the current in the battery is 2.00 mA. Find the resistances (a) R1, (b) R2, and (c) R3. Figure P18.14

(a)

To determine
The resistance R1 .

Explanation

Given info: Terminal voltage of the battery ( ε ) is 6.00 V.

Explanation:

The following diagram shows the circuit when the switch is open.

From the diagram,

εI0=R1+R2+R3

• I0 is the current in the battery when the switch is open.

Substitute 1.00 mA for I0 and 6.00 V for ε .

R1+R2+R3=6.00V1.00mAR1+R2+R3=6.00V1.00×103A

R1+R2+R3=6.00kΩ (I)

The following diagram shows the circuit when the switch is closed in position a.

From the diagram,

εIa=R1+(R2)(R2)R2+R2+R3=R1+R22+R3

• Ia is the current in the battery when the switch is closed in position a.

Substitute 1.20 mA for Ia and 6.00 V for ε .

R1+R22+R3=6.00V1.20mAR1+R22+R3=6.00V1

(b)

To determine
The resistance R2 .

(c)

To determine
The resistance R3 .

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