For the circuit shown in the following figure both meters are idealized, the bat
ID: 1280595 • Letter: F
Question
For the circuit shown in the following figure both meters are idealized, the battery has no appreciable internal resistance, and the ammeter reads 1.35 A. (Let R1 = 41.0 ?, R2 = 21.0 ?, and R3 = R4 = 11 ?.)
(a) What does the voltmeter read?
V
(b) What is the emf ? of the battery?
V
Explanation / Answer
The ammeter reads I = 1.35 A
The potential drop across the 21 ohm resistor is V25 = IR
= (1.35 A)(21 ohm)
= 28.35 V
The resistors 21 ohm, 11 ohm and (10 ohm + 11 ohm)= 21 ohm are parallel, so the potential across each will be same.
The current through 11 ohm resistor is I15 = (28.35 V)/(11 ohm)
= 2.57 A
The current through (10 ohm + 11 ohm) that is 21 ohm is I25 = (28.35 V)/(21 ohm)
= 1.35 A
Since the resistor 41 ohm is connected in series to the loop, the current through 41 ohm must be the current through the loop.
I45 = Iloop = I + I15 + I25
= 5.27 A
The potential difference across 41 ohm resistor is
V45 = ( 5.27 A)(41 ohm)
= 216.07 V
Thus, the voltmeter reads 216.07 V
Round off result to 3 significant digits, 216 V
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Let V be the voltage of the battery
Then V = IcReq
Ic is the current in the circuit = 5.27 A
Req is the equivalent resistance of the circuit
The equivalent resistance of the three resistors in the loop is
1/R' = 1/21 ohm + 1/11 ohm + 1/11 ohm
R' = 4.358 ohm
Req = 41 ohm + 35 ohm + 4.358 ohm
= 80.35 ohm
Therefore the emf of the battery is
V = (5.27 A)( 80.35 ohm)
= 423.48 V
Round off result to 3 significant digits, V= 423 V