Question

Two cells of 3V each are connected in parallel. An external resistance of 0.5 Ω is connected is series to the junction of two parallel resistors of 4 Ω and 2Ω and then to common terminal of battery through each resistor as shown in figure. What is the current flowing through 4Ω resistor?
A. 0.25 A
B. 0.55A
C. 0.35 A
D. 1.50A

Answer 1

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

Given:

• Two 3 V cells connected in parallel
• 0.5 Ω resistor connected in series
• 4 Ω and 2 Ω resistors connected in parallel

$\rule{200}2$

To Prove:

• Current flowing through 4 Ω resistor

$\rule{200}2$

Solution:

$\tt{\red{R_p=\dfrac{1}{R_1}+\dfrac{1}{R_2}}}$

$\mapsto\tt{R_p=\dfrac{1}{4}+\dfrac{1}{2}}$

$\mapsto\tt{R_p=\dfrac{1+2}{4}}$

$\mapsto{\tt{R_p=\dfrac{3}{4}}}$

$\mapsto{\green{\tt{R=\dfrac{4}{3}=1.33\:\Omega}}}$

$\rule{200}1$

$\tt{\red{Total\:resistance\:=R_s+R_p}}$

$\mapsto\tt{Total\:resistance\:=0.5+1.33}$

$\mapsto\tt{\green{Total\:resistance\:=1.83\:\Omega}}$

$\rule{200}1$

Finding the current in the circuit.

$\tt{\red{V=IR}}$

$\mapsto\tt{\red{I=\dfrac{V}{R}}}$

$\mapsto\tt{I=\dfrac{3}{1.83}}$

$\mapsto\tt{I=\dfrac{300}{183}}$

$\mapsto\tt{\green{I=1.6393\:A}}$

$\rule{200}1$

Finding potential difference in 0.5 Ω resistor.

$\tt{\red{V=IR}}$

$\mapsto\tt{V=1.6393 \times 0.5}$

$\mapsto\tt{\green{V=0.819\:V}}$

$\rule{200}1$

Finding potential difference in 4 Ω resistor.

$\tt{\red{V=Total\:voltage-Voltage\:in\:0.5\:\Omega\: resistor}}$

$\mapsto\tt{V=3-0.819}$

$\mapsto\tt{\green{V=2.181\:V}}$

$\rule{200}1$

Now, finding current flowing from 4 Ω resistor.

$\tt{\red{V=IR}}$

$\mapsto\tt{\red{I=\dfrac{V}{R}}}$

$\mapsto\tt{I=\dfrac{2.181}{4}}$

$\mapsto\tt{I=\dfrac{2181}{4000}}$

$\mapsto\tt{\green{I=0.545\:A=0.55\:A}}$

So, Option B) 0.55 A is the correct answer.

$\rule{200}4$

Answer 2

AnswEr :

• Option (B) is correct

To finD

Current flowing through the 4Ω resistor

Firstly,we would need to find the total resistance in the circuit

$\rule{300}{1}$

Total resistance in the parallel connection of 4Ω and 2Ω

$\sf \dfrac{1}{R} = \dfrac{1}{2} + \dfrac{1}{4} \\ \\ \leadsto \sf \dfrac{1}{R} = \dfrac{1 + 2}{4} \\ \\ \leadsto \boxed{\boxed{\sf R = \dfrac{4}{3} \Omega }}$

$\rule{300}{1}$

Total resistance in the circuit (series connection)

$\sf R = 0.5 + \dfrac{4}{3} \\ \\ \leadsto \sf R = \dfrac{1}{2} + \dfrac{4}{3} \\ \\ \leadsto \boxed{\boxed{\sf R = \dfrac{11}{6} \Omega}}$

$\rule{300}{1}$

Current in the circuit

From Ohm's Law,

$\star \boxed{\boxed{\tt V = IR}}$

$\tt \longrightarrow 3 = I \times \dfrac{11}{6} \\ \\ \tt \longrightarrow I = \dfrac{3 \times 6}{11} \\ \\ \large{\longrightarrow \boxed{\boxed{\tt I = \dfrac{18}{11} \ A}} }$

$\rule{300}{1}$

Potential Difference in 4Ω and 2Ω resistor

From Ohm's Law,

$\tt V = \dfrac{18}{11} \times \dfrac{4}{3} \\ \\ \large{\dashrightarrow \boxed{\boxed{ \tt V = \dfrac{24}{11} \ volts }}}$

$\rule{300}{1}$

Current in 4Ω resistor

From Ohm's Law,

$\tt I = \dfrac{V}{R} \\ \\ \longrightarrow \tt I = \dfrac{24}{4 \times 11} \\ \\ \large{\longrightarrow \boxed{\boxed{\tt I = 0.54 \ A}}}$

$\rule{300}{1}$

NOTE

• In a series connection,the current remains the same

• In a parallel connection,the potential difference remains the same

Ohm's Law :

• Potential Difference across the terminals of a conductor is directly proportional to the current flowing through it

V ∞ I

V = IR (R is resistance)

If there are n no.of resistances,

• Series Combination :

$\sf {R_{s} =R_1 + R_2 + ........... + R_n}$

• Parallel Combination :

$\sf {R_{p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + ................ + \dfrac{1}{R_n}}$

$\rule{300}{1}$

$\rule{300}{1}$