Question

(a)Three resistors 1 Ω,2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?

(b) If the combination is connected to a battery of emf 12V and negligible internal resistance, obtain the potential drop across each resistor.

Answer 1

Step 1: Given that:

Resistance($R_1$) = 1Ω

Resistance($R_2$) = 2Ω

Resistance($R_3$) = 3Ω

Emf of the battery($\in$)= 12V

Step 2: a) Calculation of total resistance of the combination of resistances in series:

The equivalent resistance of the combination of resistances in series is given by:

$R_{eq}=R_1+R_2{+R}_3$........

Thus,

The equivalent resistance of the given series combination of resistances will be

$R_{eq}=1\text{Ω}+2\text{Ω}+3\text{Ω}$

$R_{eq}=6\text{Ω}$

Step 3: b) Calculation of potential drop across each resistance:

In series combination, the current in each resistance is same whereas the potential drops across each resistance as shown in the diagram.

Using Ohm's law

$i=\frac{\in }{R_{eq}}$ (When internal resistance is zero.)

$i=\frac{12V}{6\text{Ω}}$

$i=2A$

Now,

Let, $V_1$ , $V_2$ and $V_3$ be the potential drop across resistances $R_1$ , $R_2$ and $R_3$respectively.

Thus, using Ohm's law

The potential drop across resistance $1\text{Ω}$ , is given by

$V_1=2A\times 1\text{Ω}$

$V_1=2V$

The potential drop across resistance $2\text{Ω}$ is given by;

$V_2=2A\times 2\text{Ω}$

$V_2=4V$

And, the potential drop across resistance $3\text{Ω}$ is given by;

$V_3=2A\times 3\text{Ω}$

$V_3=6V$

Thus,

a) The equivalent resistance of the combination is $6\text{Ω}$ .

b) The potential drop across given resistances$1\text{Ω},2\text{Ω}and3\text{Ω}$ are$2V,4Vand6V$ respectively.