The diagram shows a circuit. Choose the incor | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) First,simplify the circuit. The two resistors on the left ($6 \, \Omega$ and $3 \, \Omega$) are in parallel,and the two resistors on the right ($3

Vedclass · Accepted Answer

$i_3 = \frac{5}{3} \, A$

Vedclass · Answer

(B) First,simplify the circuit. The two resistors on the left ($6 \, \Omega$ and $3 \, \Omega$) are in parallel,and the two resistors on the right ($3 \, \Omega$ and $6 \, \Omega$) are in parallel. These two parallel combinations are in series with each other.Equivalent resistance of the left part: $R_L = \frac{6 	imes 3}{6 + 3} = \frac{18}{9} = 2 \, \Omega$.Equivalent resistance of the right part: $R_R = \frac{3 	imes 6}{3 + 6} = \frac{18}{9} = 2 \, \Omega$.Total equivalent resistance: $R_{eq} = R_L + R_R = 2 + 2 = 4 \, \Omega$.Total current $i_4 = \frac{V}{R_{eq}} = \frac{20}{4} = 5 \, A$.Using the current divider rule for the left side:$i_1 = i_4 	imes \frac{3}{6 + 3} = 5 	imes \frac{3}{9} = \frac{5}{3} \, A$.$i_2 = i_4 	imes \frac{6}{6 + 3} = 5 	imes \frac{6}{9} = \frac{10}{3} \, A$.For the right side,the current $i_4$ splits into $i_3$ and another branch current. By symmetry,$i_3 = i_4 	imes \frac{6}{3 + 6} = 5 	imes \frac{6}{9} = \frac{10}{3} \, A$.Comparing with the options,statement $B$ $(i_3 = \frac{5}{3} \, A)$ is incorrect.

The diagram shows a circuit. Choose the incorrect statement.

Similar Questions

Two resistors $400 \, \Omega$ and $800 \, \Omega$ are connected in series across a $6 \, V$ battery. The potential difference measured by a voltmeter of $10 \, k \Omega$ connected across the $400 \, \Omega$ resistor is close to $.... \, V$.

In the circuit shown,the resistance of the voltmeter is $10,000 \, \Omega$ and that of the ammeter is $20 \, \Omega$. The ammeter reading is $0.10 \, A$ and the voltmeter reading is $12 \, V$. Then $R$ is equal to ............. $\Omega$.

The ammeter $A$ reads $2\, A$ and the voltmeter $V$ reads $20\, V$. The value of resistance $R$ is (Assuming finite resistances of ammeter and voltmeter).

$A$ $DC$ supply of $160 \ V$ is used to charge a battery of $EMF$ $10 \ V$ and internal resistance $1 \ \Omega$ by connecting a series resistance of $24 \ \Omega$. The terminal voltage of the battery during charging is (in $V$)

In the given circuit,the current in $8 \Omega$ resistance is $1.5 \text{ A}$. The total current $(I)$ flowing in the circuit is (in $\text{ A}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module