Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Two identical cells each of emf are connected in parallel across a parallel combina

Two identical cells each of emf \mathrm{1.5\: V} are connected in parallel across a parallel combination of two resistors each of resistance \mathrm{20\, \Omega }. A voltmeter connected in the circuit measures \mathrm{1.2\, V }. The internal resistance of each cell is :

Option: 1

\mathrm{2.5\, \Omega}


Option: 2

\mathrm{4\, \Omega}


Option: 3

\mathrm{5\, \Omega}


Option: 4

\mathrm{10\, \Omega}


Answers (1)

best_answer


Simplifying the above circuit,
We get,


For cells in parallel combination,
\mathrm{E_{eq}= \frac{\frac{E_{1}}{r_{1}}+\frac{E_{2}}{r_{2}}}{\frac{1}{r_{1}}+\frac{1}{r_{2}}}=\frac{\frac{E}{r}+\frac{E}{r}}{\frac{1}{r}+\frac{1}{r}} }
         \mathrm{= E}
\mathrm{\frac{1}{r_{e q}}=\frac{1}{r}+\frac{1}{r}\, \Rightarrow\, r_{e q}=\frac{r}{2}}

\mathrm{\because \; E= 1.5\, V\; \; \left ( given \right )}
\mathrm{I=\frac{E}{\frac{r}{2}+10}=\frac{(1. 5)}{r+20}=\frac{3}{r+20}\; \rightarrow (1)}

\mathrm{E-I(r / 2)=I(10)=V \text { (Reading of volume) }}
\mathrm{I(10)=1.2 \mathrm{~V}}
I = 0.12 A
\mathrm{0.12 A=\frac{3}{r+20}}
\mathrm{0.12 \, r=0.6}

\mathrm{r=5 \Omega}
The correct option is (3)
 

Posted by

Shailly goel

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: BTech ECE cut-offs for female candidates go up at top NITs
anam-khan

JEE Main 2025 Session 2 City Slip (OUT) Live: Admit card likely on March 29 at jeemain.nta.nic.in; cut-offs
anam-khan

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads

Latest Question