Non-Ideal Voltmeter & Non-Ideal Ammeter

ElecCircNonIdealVolAmp
A student built an electric circuit that consists of an ideal battery, a voltmeter with resistance RV, an ammeter with resistance RA and a resistor R. She connected all elements in the circuit as shown in Fig. (1). The voltmeter showed a reading of 1V and the ammeter showed a current of 1A. Then, another student switched between the voltmeter and ammeter Fig. (2). As a result, the voltmeter showed the reading of 2V and the ammeter showed the current of 0.5A. What is the resistance of the resistor?

Solution.

First, let’s look at the circuit diagram in Fig. (1). The ammeter showed the current of 1A and its resistance is RA. According to the Ohm’s law, the voltage on the ammeter is equal to 1*RA.
The Kirchhoff’s Voltage law states that for a closed loop series path the algebraic sum of all the voltages around any closed loop in a circuit is equal to zero. The battery is ideal, the potential difference between its terminals is equal to ε. The voltmeter showed the reading of 1V, so,

ε – 1*RA – 1 = 0

or

ε = RA + 1   (1)

Now we can look at the circuit diagram in Fig. (2). The voltmeter showed the reading of 2V. According to the Ohm’s law, the voltage on the ammeter is equal to 0.5*RA. Then,

ε – 2 – 0.5*RA = 0

or

ε = 2 + 0.5*RA   (2)

The left-hand side of equation (1) and the left-hand side of equation (2) is equal to one another. So we can simply set the right-hand sides of the two equations equal to one another, then solve the resulting equation.

RA + 1 = 2 + 0.5*RA

RA = 2Ω.

We can find ε by substituting the value of RA into the equation (1)

ε = 2 + 1 = 3V.

Now we can use Kirchhoff’s current law which states that the total current entering a circuit junction is exactly equal to the total current leaving the same junction.
 
First, we look at the circuit diagram in Fig. (1). The current entering the junction is measured by ammeter and equals 1A. There are two currents leaving the junction. According to the Ohm’s law, one branch current is 1/R and the other is 1/RV. Thereby,

1 = 1/R + 1/RV   (3)

Now we can look at the circuit diagram in Fig. (2). According to the Ohm’s law, the current entering the junction is 2/RV. The total voltage or potential difference of the battery is 3V and voltmeter showed the reading of 2V. So, the voltage on the ammeter or resistor is 1V. There are two currents leaving the junction. One branch current is 1/R and the other is measured by ammeter and equals 0.5A. Thereby,

2/RV = 1/R + 0.5   (4)

We can solve the system of equations (3) and (4).

We rewrite the equation (3) as

1/R = 1 – 1/RV   (5).

Also, we rewrite the equation (4) as

1/R = 2/RV – 0.5   (6).

The left-hand side of equation (5) and the left-hand side of equation (6) is equal to one another. So we can simply set the right-hand sides of the two equations equal to one another, then solve the resulting equation.

1 – 1/RV = 2/RV – 0.5

1.5 = 3/RV

RV = 2Ω.

We substitute the value of RV into the equation (3) and solve for R.

1 = 1/R + 1/2

R = 2Ω.

It is important to mention that the ammeter and voltmeter in this problem are very bad. They both have resistance that they shouldn’t have. The voltmeter should have very large resistance and the ammeter should have very small resistance.

 
 
Thank you!
 
 

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