### Resistance in Parallel:

When resistors are placed in parallel, they are arranged side by side with their ends joined together. With the arrangement, the potential difference across the resistors is the same.

The positive terminals of all the resistors are connected to the same positive terminal and the negative terminals of the resistors are connected to the same negative terminal.

The equipment parallel resistance R_{total} is found using the expression,

I = I_{1} + I_{2} + I_{3 ____________________ }**(1)**

Since the current flowing through each resistor is

I_{1} = \( \frac{V}{R_1}\), I_{2} = \( \frac{V}{R_2}\), I_{3} = \( \frac{V}{R_3}\) __________________ **(2)**

Substituting these equations **(2)** in **(1)**

∴ I = \( \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3} =\frac{V}{R_{total}}\)

= \( \scriptsize V \left ( \normalsize \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right) = \normalsize \frac{V}{R_{total}}\)

**cross multiply**

∴ \(\frac{1}{R_{total}} = \frac{1}{R_1}+ \frac{1}{R_2} + \frac{1}{R_3}\)

For two resistors in parallel

Total Resistance = \( \frac {R_1 R_2}{R_1 + R_2} \)

### Example 1:

Three resistors 6 Ω, 8 Ω and 10 Ω are connected in parallel. Calculate the total resistance of the circuit.

**Solution:**

**Data given in the question:**

R_{1} = 6 Ω, R_{2} = 8 Ω, R_{3} = 10 Ω

**Formula:** \(\frac{1}{R_{total}} = \frac{1}{R_1} \: + \: \frac{1}{R_2} \: + \: \frac{1}{R_3}\)

**Substitute the values into the equation:**

⇒ \(\frac{1}{R_{total}} = \frac{1}{6}+ \frac{1}{8} + \frac{1}{10}\)

L.C.M = 120

⇒ \(\frac{1}{R_{total}} = \frac{20\:+\:15\:+\:10}{120}\)

⇒ \(\frac{1}{R_{total}} = \frac{45}{120}\)

∴ \( \scriptsize R_{total} = \normalsize \frac{120}{45}\)

⇒ \( \scriptsize R_{total} = 2.67 \Omega\)

### Example 2:

Calculate the total resistance of the circuit below.

**Solution:**

**Data given in the question:**

R_{1} = 2 Ω, R_{2} = 6 Ω, R_{3} = 3 Ω

**Formula:** \(\frac{1}{R_{total}} = \frac{1}{R_1} \: + \: \frac{1}{R_2} \: + \: \frac{1}{R_3}\)

**Substitute the values into the equation:**

⇒ \(\frac{1}{R_{total}} = \frac{1}{2}+ \frac{1}{6} + \frac{1}{3}\)

L.C.M = 12

⇒ \(\frac{1}{R_{total}} = \frac{6\:+\:2\:+\:4}{12}\)

⇒ \(\frac{1}{R_{total}} = \frac{12}{12}\)

⇒ \(\frac{1}{R_{total}} = \frac{1}{1}\)

∴ \( \scriptsize R_{total} = 1 \Omega\)

### Example 3:

Find the total resistance in the given circuit below.

**Solution:**

R_{total} = Total Resistance in series + Total Resistance in parallel

Total Resistance in series, R_{series} = 2 Ω + 3 Ω = 5 Ω

Total Resistance in parallel ⇒ \(\frac{1}{R_{parallel}} = \frac{1}{5}\: +\: \frac{1}{5} \\ = \frac{1}{R_{parallel}} = \frac{2}{5} \\ \scriptsize R_{parallel} = \normalsize \frac{5}{2} \\ \scriptsize R_{parallel} = 2 \frac{1}{2} \Omega\)

The resultant resistive circuit now looks something like this:

We can see that the two remaining resistances are connected together in a “SERIES” combination and again they can be added together.

R_{total} = R_{series} + R_{parallel}

= \( \scriptsize 5\: \Omega \: + \: 2 \frac{1}{2}\: \Omega \)

R_{total} = \( \scriptsize 7 \frac{1}{2}\: \Omega \)

### Example 3:

Calculate the total current ( IT ) taken from the 12V supply in the following circuit.

**Data given in the question:**

R_{1} = 6 Ω, R_{2} = 8 Ω, R_{3} = 4 Ω, R_{4} = 12 Ω

Voltage = 12V, I_{T} = ?

R_{2} and R_{3} are connected together in **series** so we can add them together to produce the aggregate resistance.

R_{2} + R_{3} = 8Ω + 4Ω = 12Ω

We can replace both resistor R_{2} and R_{3} above with a single resistor R_{x} of resistance value 12Ω

From this new arrangement, we can now see resistor R_{x} is parallel with the resistor R_{4}.

We can calculate the total resistance of the parallel circuits using the formula;

**Formula:** \(\frac{1}{R_{total}} = \frac{1}{R_x} \: + \: \frac{1}{R_4} \)

⇒ \(\frac{1}{R_{total}} = \frac{1}{12} \: + \: \frac{1}{12} \)

⇒ \(\frac{1}{R_{total}} = \frac{2}{12} \)

⇒ \(\frac{1}{R_{total}} = \frac{1}{6} \)

∴ \( \scriptsize R_{total} = 6 \: \Omega \)

The resultant resistive circuit now looks something like this:

From the diagram the remaining tow resistors are connected in series, so we add them to get the total resistance of the circuit

R_{total} = R_{1} + R_{y}

R_{total} = 6Ω + 6Ω

R_{total} = 12Ω

By using Ohm’s Law, the value of the current ( I ) flowing around the circuit is calculated as:

I = \( \frac{V}{R}\)

V = 12V, R = 12Ω

I = \( \frac{12}{12}\)

I = 1 Ampere (A)

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