Resistors In Series And Parallel Combinations

What is a Series Grouping of Resistance?

In this case, the Potential drop is different across each resistor and the Current is the same

.

$
R_{e q}=R_1+R_2+R_3+\cdots+R_n
$
$R_{e q}={ }_{\text {Equivalent Resistance }}$
For n identical resistance: $R_{\text {eq }}=n R$
$
V^{\prime}=\frac{V}{n}
$

What is Parallel Grouping of Resistance?

In this case, the Potential is the Same across each resistor and the current is different

$
\frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n}
$

If two resistances are in Parallel:
$
R_{e q}=\frac{R_1 R_2}{R_1+R_2}
$

Current through any resistance:

$
i^{\prime}=i\left(\frac{\text { Resistance of opposite Branch }}{\text { total Resistance }}\right)
$

The required current of the first branch
$
i_1=i\left(\frac{R_2}{R_2+R_2}\right)
$

The required current of the second branch
$
i_2=i\left(\frac{R_1}{R_1+R_2}\right)
$

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Solved Examples Based on Resistors In Series And Parallel Combinations

Example 1: The total current (in amperes) supplied to the circuit by the battery is

1) 4

2) 2

3) 1

4) 6

Solution:

The equivalent circuits are shown below :

$I=\frac{6}{1.5}=4 \mathrm{~A}$

Hence, the answer is option (1).

Example 2: In the figure shown, what is the current (in ampere) drawn from the battery? you are given:

$R_1=15 \Omega, R_2=10 \Omega, R_3=20 \Omega, R_4=5 \Omega, R_5=25 \Omega, R_6=30 \Omega, E=15 \mathrm{~V}$

1) 13/24

2) 7/18

3) 9/32

4) 20/3

Solution:

Series Grouping

Potential - Different

Current - Same

Parallel Grouping -

Potential - Same

Current - Different

$\begin{aligned} & I=\frac{V}{R_{\mathrm{cq}}} \\ & R_{\text {eq }}=R_1+R_6+\frac{1}{\frac{1}{R_2}+\frac{1}{R_3+R_1+R_5}} \\ & =15+30+\frac{1}{\frac{1}{10}+\frac{1}{20+5+25}} \\ & =15+30+\frac{25}{3} \\ & =\frac{135+25}{3} \\ & =\frac{160}{3} \\ & I=\frac{15}{\frac{160}{3}}=\frac{45}{160}=\frac{9}{32}\end{aligned}$

Hence, the answer is option (3).

Example 3: A 3-volt battery with negligible internal resistance is connected in a circuit as shown in the figure. The current I (in Amperes) in the circuit will be

1) 1.5

2) 1

3) 2

4) 0.33

Solution:

$\begin{aligned} & \operatorname{Req}=\frac{(3+3) \times 3}{(3+3)+3}=\frac{18}{9}=2 \Omega \\ & I=\frac{V}{R}=\frac{3}{2}=1.5 \mathrm{~A}\end{aligned}$

Example 4: The total current (in amperes) supplied to the circuit by the battery is:

1) 4

2) 2

3) 1

4) 6

Solution:

The equivalent circuits are shown below :

$I=\frac{6}{1.5}=4 \mathrm{~A}$

Hence, the answer is option (1).

Example 5: In the given circuit, an ideal voltmeter connected across the $10 \Omega$ resistance reads 2 V . The internal resistance r (in $\Omega$ ), of each cell, is :

1) 0.5

2) 1

3) 1.5

4) 0

Solution:

In series Grouping

$
R_{c q}=R_1+R_2+R_3+\cdots+R_n
$
wherein
$R_{c q}-$ Equivalent Resistance

In parallel Grouping
$
\frac{1}{R_{c q}}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n}
$

So,

$\begin{aligned} \mathrm{V} & =6 \times \mathrm{i}=2 \\ i & =\frac{1}{3} A \\ i & =\frac{2 E}{2 r+2+6} \\ & =\frac{2 \times 1.5}{2 r+8} \\ & =\frac{1}{3} \\ \Rightarrow & \Rightarrow 9=2 r+8 \\ \Rightarrow & \Rightarrow r=0.5 \Omega\end{aligned}$

Summary

As per a series combination, resistors are linked from one end to the next; the overall resistance is the sum of all individual resistances hence equal currents flowing through them with a similar voltage drop across. In the case of a parallel combination, on the other hand, resistors are connected using two common points; one can see that total resistance is always less than any one of the individual resistors leading to the same voltage drop across.