In the circuit at right the switch is initially open. We want to measure the pot

Question

In the circuit at right the switch is initially open. We want to measure the potential across the unknown battery but our voltmeter is broken. We know the resistance of resistor R-, to good precision from previous measurements. We do not know the resistance of R2 to sufficient precision to properly measure the potential across the battery. So first we'll measure the resistance of R2. What is the potential difference across R2 when the switch is open? With the switch open the ammeter reads 1. 00 a. What is the resistance of resistor R2 to an appropriate precision? When the switch is closed the ammeter reads 1. 57 a. What is the potential difference across resistor R2 now? What is the potential difference across the unknown battery?

Explanation / Answer

A) when switch is open

i=6/(4+r2)

So

Potential across r2=6-4i=6-24/(4+r2)=6r2/(4+r2)

B)i=1=6/(4+r2)

So

r2=2ohms

C)When switch is closed we have these equations then

from right loop we have

e=1.57*r2

So

e=3.14 Volts

D)So potential difference across the unknown battery=e=3.14 Volts

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