Question

(1) Three resistors $1\Omega ,2\Omega ,\text{and}3\Omega$ are combined in series.
$A)$What is the total resistance of the combination?

(2) Three resistors $1\Omega ,2\Omega ,\text{and}3\Omega$ are combined in series.
(B)If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Answer 1

(1) $Step1:$ Draw the diagram
Step 2: Find the total series resistance
Formula used:$R_S=R_1+R_2+R_3$

Three resistors of resistances $1\Omega ,2\Omega ,$ and $3\Omega$ are

combined in series. Total resistance of the combination is

given by the algebraic sum of individual resistances,$i.e.$
$R_S=R_1+R_2+R_3$

$R_S=1+2+3=6\Omega$

$2)$ Step 1: Find the Potential drop across $1\Omega$
Formula used: $V=IR$

Current flowing through the circuit $=I$

Emf of the battery,$E=12V$

Total resistance of the circuit, $R=6\Omega$

The relation for current using Ohm's law is,
$I=\frac{E}{R}$
$=\frac{12}{6}=2A$

Potential drop across $1\Omega$ resistor $\left(V_1\right)$

From Ohm's law, the value of $V_1$ can be obtained as

$V_1=2\times 1=2V$ …$\left(i\right)$

Step 2: Find the Potential drop across $2\Omega$

Potential drop across $2\Omega$ resistor $\left(V_2\right)$

Again, from Ohm's law, the value of $V_2$ can be obtained as

$V_2=2\times 2=4V$ …$\left(ii\right)$

Step 3:Find the Potential drop across $3\Omega .$

Potential drop across $3\Omega$ resistor $\left(V_3\right)$

Again, from Ohm's law, the value of $V_3$ can be obtained as
$V_3=2\times 3=6V$ …$\left(iii\right)$

Final Answer: Total resistance $=6\Omega$

Potential drop across $1\Omega$ resistor $\left(V_1\right)=2V$

Potential drop across$2\Omega$ resistor $\left(V_2\right)=4V$

Potential drop across $3\Omega$ resistor $\left(V_3\right)=6V$