Current Electricity — Questions & Answers

Sample questions from Current Electricity

1. Q1. A hollow cylindrical conductor of resistivity 2.0 x 10^-8 ohm m has inner radius 1.0 mm, outer radius 2.0 mm and length 0.50 m. Current flows ALONG its length. Its resistance is closest to:

   • A.1.1 x 10^-3 ohm✓
   • B.3.2 x 10^-3 ohm
   • C.1.1 x 10^-6 ohm
   • D.8.0 x 10^-3 ohm
   Solution

   The conducting cross-section is the annulus A = pi(b^2 - a^2) = pi(4 - 1) x 10^-6 = 9.42 x 10^-6 m^2. Then R = rho L/A = (2.0 x 10^-8 x 0.50)/(9.42 x 10^-6) = 1.06 x 10^-3 ohm. Using the full outer disc instead of the annulus underestimates R by a factor 3/4.

2. Q2. For the network of the previous type with arms AB = 3 ohm, BC = 6 ohm, AD = 6 ohm, DC = 3 ohm and a 4 ohm bridging resistor between B and D, the equivalent resistance between A and C is:

   • A.4.5 ohm
   • B.72/17 ohm, about 4.24 ohm✓
   • C.9 ohm
   • D.2.25 ohm
   Solution

   Because the bridge is unbalanced, the 4 ohm branch cannot be deleted. With a 9 V source the nodal solution gives a total supply current of 17/8 A, so R_eq = 9/(17/8) = 72/17 ohm, close to 4.24 ohm. Ignoring the bridging branch would wrongly give the two 9 ohm paths in parallel, 4.5 ohm.

3. Q3. Two cells of EMF 12 V (internal resistance 2 ohm) and 6 V (internal resistance 3 ohm) are joined in parallel with like terminals together, and the combination feeds a 4 ohm external resistor. The current supplied to the external resistor is:

   • A.2.31 A
   • B.3.00 A
   • C.0.46 A
   • D.1.85 A✓
   Solution

   E_eq = (12/2 + 6/3)/(1/2 + 1/3) = 8/(5/6) = 9.6 V and r_eq = 1/(5/6) = 1.2 ohm. Then I = 9.6/(1.2 + 4) = 1.85 A. Notably the 6 V cell carries 0.46 A in the REVERSE sense, i.e. it is being charged by the stronger cell.

4. Q4. A cell of EMF 2V and internal resistance 0.5 ohm is connected to an external resistance that draws maximum power from the cell. The value of this external resistance and the maximum power delivered are:

   • A.0.5 ohm, 2W✓
   • B.0.5 ohm, 4W
   • C.1 ohm, 2W
   • D.0.25 ohm, 4W
   Solution

   Maximum power transfer occurs at R=r=0.5 ohm. At this point, I = E/(R+r) = 2/1 = 2A, and power delivered to R: P=I^2 R = 4 x 0.5 = 2W. A student who picks R=r correctly but miscalculates the power (perhaps using the full EMF x current, 2x2=4W, forgetting only half the power goes to R and half is lost internally) picks B.

5. Q5. Two 10 kΩ resistors are connected in series across a 30 V ideal source. A voltmeter of resistance 10 kΩ is connected across one of them. The voltmeter reading is:

   • A.15 V
   • B.10 V✓
   • C.20 V
   • D.7.5 V
   Solution

   The voltmeter in parallel with one 10 kΩ gives 5 kΩ. Total = 15 kΩ, current = 2 mA, so the reading = 2 mA × 5 kΩ = 10 V, well below the true 15 V. This loading error is why a voltmeter needs a resistance much larger than the circuit resistance.

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