Currents through Parallel Resistances

Refer to the schematic diagram of following figure. The resistances are called Rn. The total parallel resistance in the circuit is R. The battery voltage is E. The current in any particular branch n, containing resistance Rn, is measured by ammeter A and is called In. The sum of all the currents In is equal to the total current, I, drawn from the battery. The current is divided up in the parallel circuit in a manner similar to the way that voltage is divided up in a series circuit.


Conventional Current

Have you noticed that the direction of current flow in above figure is portrayed as outward from the positive battery terminal? Don’t electrons, which are the actual charge carriers in a wire, flow out of the minus terminal of a battery? Yes, that’s true; but scientists consider theoretical current, more often called conventional current (because it is defined by convention), to flow from positive to negative voltage points, rather than from negative to positive.

Problem 1

Suppose that the battery in above figure delivers 12 V. Further suppose that there are 12 resistors, each with a value of 120 Ω in the parallel circuit. What is the total current, I, drawn from the battery?
First, find the total resistance. This is easy, because all the resistors have the same value. Just divide Rn = 120 by 12 to get R = 10 Ω. Then the current can be found by Ohm’s Law: I = E/R = 12/10 = 1.2 A.

Problem 2

In the circuit of above figure, what does the ammeter say?
This involves finding the current in any given branch. The voltage is 12 V across every branch, and Rn = 120 Ω. Therefore In, the ammeter reading, is found by Ohm’s Law: In = E/Rn = 12/120 = 0.10 A.

Because this is a parallel circuit, all of the branch currents In should add up to get the total current, I. There are 12 identical branches, each carrying 0.10 A; therefore the total current is 0.10 × 12 = 1.2 A. It checks out.

Problem 3

Suppose three resistors are in parallel across a battery that supplies E = 12 V. The resistances are R1 = 22 Ω, R2 = 47 Ω, and R3 = 68 Ω. These resistances carry currents I1, I2, and I3, respectively. What is the current, I3, through R3?

This problem is solved by means of Ohm’s Law as if R3 is the only resistance in the circuit. There’s no need to worry about the parallel combination. The other branches do not affect I3. Thus I3 = E/R3 = 12/68 = 0.18 A.

Problem 4

What is the total current drawn by the circuit described in Problem 3?
There are two ways to go at this. One method involves finding the total resistance, R, of R1, R2, and R3 in parallel, and then calculating I based on R. Another way is to find the currents through R1, R2, and R3 individually, and then add them up.

Using the first method, first change the resistances Rn into conductances Gn. This gives G1 = 1/R1 = 1/22 = 0.04545 S, G2 = 1/R2 = 1/47 = 0.02128 S, and G3 = 1/R3 = 1/68 = 0.01471 S. Adding these gives G = 0.08144 S. The resistance is therefore R = 1/G = 1/0.08144 = 12.279 Ω. Use Ohm’s Law to find I = E/R = 12/12.279 = 0.98 A. Note that extra digits are used throughout the calculation, rounding off only at the end.

Now let’s try the other method. Find I1 = E/R1 = 12/22 = 0.5455 A, I2 = E/R2 = 12/47 = 0.2553 A, and I3 = E/R3 = 12/68 = 0.1765 A. Adding these gives I = I1 + I2 + I3 = 0.5455 + 0.2553 + 0.1765 = 0.9773 A, which rounds off to 0.98 A.