Parallel Networks

shows three resistors, R₁, R₂ and R₃ connected across each other, i.e., in parallel, across a battery source of V volts.

Figure 3.9 : Parallel resistors

In a parallel circuit:

(a) the sum of the currents I₁, I₂ and I₃ is equal to the total circuit current, I, i.e., I =I₁ +I₂ +I₃, and

(b) the source voltage, V volts, is the same across each of the resistors.

From Ohm’s law:

where R is the total circuit resistance.

Since I =I₁ +I₂ +I₃

then 

Dividing throughout by V gives:

This equation must be used when finding the total resistance R of a parallel circuit. For the special case of two resistors in parallel:

Hence, 

Example 3.6

For the circuit shown in Figure 3.10, determine (a) the reading on the ammeter, and (b) the value of resistor R₂.

Figure 3.10 : Circuit for Example 3.6

Solution

Voltage across R₁ is the same as the supply voltage V. Hence, supply voltage V = 8×5=40 V.

(a) Reading on ammeter, 

(b) Current flowing through R₂ = 11-8-2=1 A

Hence, 

Example 3.7

Two resistors, of resistance 3 Ω and 6 Ω, are connected in parallel across a battery having a voltage of 12 V. Determine (a) the total circuit resistance and (b) the current flowing in the 3 Ω resistor.

Solution

The circuit diagram is shown in Figure 3.11.

(a) The total circuit resistance R is given by:

Figure 3.11 : Circuit for Example 3.7

Hence, 

(b) Current in the 3 Ω resistance, 

Example 3.8

For the circuit shown in Figure 3.12, find (a) the value of the supply voltage V and (b) the value of current I.

Figure 3.12 : Circuit for Example 3.8

Solution

(a) Voltage across 20 Ω resistor =I₂R₂ = 3×20=60 V; hence, supply voltage V=60 V since the circuit is connected in parallel.

(b) Current 

Current I =I₁ +I₂ +I₃ and hence, I = 6+3+1=10 A

Alternatively, 

Hence, total resistance 

Current 

Example 3.9

Given four 1 Ω resistors, state how they must be connected to give an overall resistance of (a) 1/4 Ω (b) 1 Ω (c)  (d) 

Solution

(a) All four in parallel (see Figure 3.13),

Figure 3.13 : Circuit for Example 3.9(a)9

Since 

(b) Two in series, in parallel with another two in series (see Figure 3.14), since 1 Ω and 1 Ω in series gives 2 Ω, and 2 Ω in parallel with 2 Ω gives:

Figure 3.14 : Circuit for Example 3.9(b)9

(c) Three in parallel, in series with one (see Figure 3.15), since for the three in parallel,

Figure 3.15 : Circuit for Example 3.9(c)9

 in series with 1 Ω gives 

(d) Two in parallel, in series with two in series (see Figure 3.16), since for the two in parallel

Figure 3.16 : Circuit for Example 3.9(d)9

 and  1 Ω and 1 Ω in series gives 

Example 3.10

Find the equivalent resistance for the circuit shown in Figure 3.17.

Figure 3.17 : Circuit for Example 3.10

Solution

R₃, R₄ and R₅ are connected in parallel and their equivalent resistance R is given by:

Hence, 

The circuit is now equivalent to four resistors in series and the equivalent circuit resistance = 1+2.2+1.8+4=9 Ω.