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Resistors in Parallel Help

By — McGraw-Hill Professional
Updated on Sep 15, 2011

Voltage Across Parallel Resistances

Imagine now a set of ornamental light bulbs connected in parallel. This is the method used for outdoor holiday lighting or for bright indoor lighting. You know that it’s much easier to fix a parallel-wired string of holiday lights if one bulb should burn out than it is to fix a series-wired string. The failure of one bulb does not cause catastrophic system failure. In fact, it might be awhile before you notice that the bulb is dark because all the other ones will stay lit, and their brightness will not change.

In a parallel circuit, the voltage across each component is always the same and is always equal to the supply or battery voltage. The current drawn by each component depends only on the resistance of that particular device. In this sense, the components in a parallel-wired circuit work independently, as opposed to the series-wired circuit, in which they all interact.

If any branch of a parallel circuit is taken away, the conditions in the other branches remain the same. If new branches are added, assuming that the power supply can handle the load, conditions in previously existing branches are not affected.

Currents Through Parallel Resistances

Refer to the schematic diagram of Fig. 12-12. The total parallel resistance in the circuit is R . The battery voltage is E . The current in branch n , containing resistance R n , is measured by ammeter A and is called I n .

The sum of all the I n values in the circuit is equal to the total current I drawn from the source. That is, the current is divided up in the parallel circuit, similarly to the way that voltage is divided up in a series circuit.

Currents Through Parallel Resistances Practice Problem

Problem 1

Suppose that the battery in Fig. 12-12 delivers 12 V. Further suppose that there are 12 resistors, each with a value of 120 ohms in the parallel circuit.

 

Direct Current How Resistances Combine Currents Through Parallel Resistances

Fig. 12-12 . Analysis of current in a parallel dc circuit. See text for discussion.

What is the total current I drawn from the battery?

Solution 1

First, find the total resistance. This is easy because all the resistors have the same value. Divide R n = 120 by 12 to get R = 10 ohms. Then the current I is found by Ohm’s law:

I = E/R = 12/10 = 1.2 A

Problem 2

In the circuit of Fig. 12-12 , what does the ammeter A say, given the same component values as exist in the scenario of the preceding problem?

Solution 2

This involves finding the current in any given branch. The voltage is 12 V across every branch; R n = 120. Therefore, I n , the ammeter reading, is found by Ohm’s law:

 

I n = E/R n = 12/120 = 0.10 A

Let’s check to be sure all the I n values add to get the total current I . There are 12 identical branches, each carrying 0.10 A; therefore, the sum is 0.10 × 12 = 1.2 A. It checks out.

Power Distribution In Parallel Circuits

When resistances are wired in parallel, they each consume power according to the same formula, P = I 2 R . However, the current is not the same in each resistance. An easier method to find the power P n dissipated by resistor of value R n is by using the formula P n = E 2 / R n , where E is the voltage of the supply. This voltage is the same across every resistor.

In a parallel circuit, the total power consumed is equal to the sum of the wattages dissipated by the individual resistances. This is, in fact, true for any dc circuit containing resistances. Power cannot come out of nowhere, nor can it vanish.

Power Distribution In Parallel Circuits Practice Problem

Problem

A circuit contains three resistances R 1 = 22 ohms, R 2 = 47 ohms, and R 3 = 68 ohms, all in parallel across a voltage E = 3.0 V. Find the power dissipated by each resistor.

Solution

First find E 2 , the square of the supply voltage: E 2 = 3.0 × 3.0 = 9.0. Then P 1 = 9.0/22 = 0.4091 W , P 2 = 9.0/47 = 0.1915 W, and P 3 = 9.0/68 = 0.1324 W. These should be rounded off to P 1 = 0.41 W, P 2 = 0.19 W, and P 3 = 0.13 W, respectively.

Practice problems of these concepts can be found at: Direct Current Practice Test

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