Solutions to Part 21 (Electric circuits)

We now undertake the solutions for the problems in Part 21, Introducing Basic Physics: Simple electric circuits. The problem will be given then the solution.

1) A 12 V battery has an internal resistance of 2 ohms. If it is connected in series with a voltmeter and another resistance R = 4 ohms, what would the voltmeter read? What would an ammeter read placed in the same circuit?

The circuit set-up is shown for Problem (1). We first need to find the ammeter reading:

I = E/ (R + r) = 12V /(4 ohms + 2 ohms)= 12V/ 6 ohms = 2 A

Then the voltmeter (V) reading may be obtained from Ohm's law for the circuit:

V = I(R) = 2A (4 ohms) = 8V (at the position indicated)

This may also be validated by use of:

V = E - Ir = 12V - (2A)(2 ohms) = 12V - 4V = 8V

(e.g. 4V of total emf is lost through the source)


2) A Wheatstone Bridge circuit is connected as shown in Fig. 1(b). The galvanometer is found to read zero when point C is located exactly midway along a wire 1m in length (e.g. connecting A and B in the diagram). A known resistance coil R is used which is made of copper (rho = 1.72 x 10^-8 ohm-m) and is 50 m long, wound tightly in a coil.

a) If the cross sectional area A = πr^2 and r = 0.001m, find the value of R(x).


The experimental circuit is shown for Problem 2, with L1 and L2 denoting the respective lengths.

We first need to obtain the known resistance, but this must be done using the resistivity of the wire that's given (rho = 1.72 x 10^-8 ohm-m) in conjunction with the resistance as a function of resistivity eqn.

R = rho(L)/A = rho(L)/ πr^2

R = (rho = 1.72 x 10^-8 ohm-m)(50m)/(π x.001m^2) = 0.27 ohms


And, from the Wheatstone Bridge set up:

R(x)/ R = L1/L2

But since: L1 = L2 = 50 cm, then:

R(x) = R (L1/L2) = R (1) = 0.27 ohms



b) If a new resistor R made up of 100m length of the same copper wire is then used, then how must the lengths L1 and L2 change to achieve a galvanometer reading of zero?

We assume the only change made is to R, and R(x) is still 0.27 ohms. Then only the lengths L1, L2 will vary.

The new known resistance, call it R' = rho(2L)/A = 2R = 0.54 ohms

(since 100m = 2(50m))

Then:

R(x)/ R' = L1/ L2 = 0.27 ohms/ 0.54 ohms = 0.5

So: L1 = 0.5 (L2) or L2 = 2L1

The total length is 1m, so:

L1 + L2 = 100 cm

substituting for L2 (2L1):

L1 + 2L1 = 3L1 = 100 cm and L1 = 100cm/ 3 = 33.3 cm

So: L2 = 100 cm - L1 = 100cm - 33.3 cm = 66.7 cm