In the given figure R_1=10 , R_2=8 , R_3=4 and R_4=8 . Battery is ideal with emf 12 ~V . Equivalent resistant of the circuit and current supplied by battery are respectively :

Correct answer: B. 12 and 1 ~A Here R_2, R_3, R_4 are in parallel aligned & 1 R_ 234 = 1 R_2 + 1 R_3 + 1 R_4 & R_ 234 =2 aligned R_ 234 is in series with R_1 so aligned & R_ eq =R_ 234 +R_1=2+10=12 & i=(12)/(12)=1 Amp aligned

JEE Main 20245 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 5 April
Shift: Morning Shift
Subject: Physics
Chapter: Current Electricity

JEE Main 2024 · 5 April · Morning Shift · Physics · Current Electricity

In the given figure R_1=10 , R_2=8 , R_3=4 and R_4=8 . Battery is...

In the given figure

\mathrm{R}_1=10 \Omega, \mathrm{R}_2=8 \Omega, \mathrm{R}_3=4 \Omega

and

\mathrm{R}_4=8 \Omega

. Battery is ideal with emf

12 \mathrm{~V}

. Equivalent resistant of the circuit and current supplied by battery are respectively : In the given figure R_1=10 , R_2=8 , R_3=4 and R_4=8 . Battery is ideal with emf 12 ~V . Equivalent resistant of the

In the given figure R_1=10 , R_2=8 , R_3=4 and R_4=8 . Battery is ideal with emf 12 ~V . Equivalent resistant of the

Options

A $10.5 \Omega$ and $1.14 \mathrm{~A}$
B $12 \Omega$ and $1 \mathrm{~A}$
C $10.5 \Omega$ and $1 \mathrm{~A}$
D $12 \Omega$ and $11.4 \mathrm{~A}$

Correct answer

B. $12 \Omega$ and $1 \mathrm{~A}$

Step-by-step solution

Here

\mathrm{R}_2, \mathrm{R}_3, \mathrm{R}_4

are in parallel

\begin{aligned} & \frac{1}{\mathrm{R}_{234}}=\frac{1}{\mathrm{R}_2}+\frac{1}{\mathrm{R}_3}+\frac{1}{\mathrm{R}_4} \\ & \mathrm{R}_{234}=2 \Omega \end{aligned}

R_{234}

is in series with

R_1

so

\begin{aligned} & \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{234}+\mathrm{R}_1=2+10=12 \Omega \\ & \mathrm{i}=\frac{12}{12}=1 \mathrm{Amp} \end{aligned}

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