Find the current in the 3,Omega resistance in | vedclass

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

Question

(C) The circuit consists of a $14\,V$ battery in series with a $5\,\Omega$ resistor,which is then connected in parallel to two branches: one with a $3

Vedclass · Accepted Answer

$\frac{4}{3}\,A$

Vedclass · Answer

(C) The circuit consists of a $14\,V$ battery in series with a $5\,\Omega$ resistor,which is then connected in parallel to two branches: one with a $3\,\Omega$ resistor and another with a $(2+4)\,\Omega = 6\,\Omega$ resistor.First,calculate the equivalent resistance of the parallel part:$R_p = \frac{3 	imes 6}{3 + 6} = \frac{18}{9} = 2\,\Omega$The total equivalent resistance of the circuit is:$R_{eq} = R_p + 5\,\Omega = 2\,\Omega + 5\,\Omega = 7\,\Omega$The total current $I$ flowing from the battery is:$I = \frac{V}{R_{eq}} = \frac{14\,V}{7\,\Omega} = 2\,A$Using the current divider rule,the current $I_1$ through the $3\,\Omega$ resistor is:$I_1 = I 	imes \left( \frac{R_{other}}{R_{3\Omega} + R_{other}} ight) = 2\,A 	imes \left( \frac{6\,\Omega}{3\,\Omega + 6\,\Omega} ight) = 2 	imes \frac{6}{9} = 2 	imes \frac{2}{3} = \frac{4}{3}\,A$

Find the current in the $3\,\Omega$ resistance in the given circuit.

Similar Questions

$A$ storage battery of $emf$ $8.0\; V$ and internal resistance $0.5\; \Omega$ is being charged by a $120\; V$ $DC$ supply using a series resistor of $15.5\; \Omega$. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

The equivalent resistance between $A$ and $B$ is ............ $\Omega$.

Three resistors having resistances $r_{1}, r_{2}$ and $r_{3}$ are connected as shown in the given circuit. The ratio $\frac{i_{3}}{i_{1}}$ of currents in terms of resistances used in the circuit is :

The equivalent resistance of the infinite network given below is:

$B_{1}$, $B_{2}$ and $B_{3}$ are three identical bulbs connected to a battery of steady $EMF$ with key $K$ closed. What happens to the brightness of the bulbs $B_{1}$ and $B_{2}$ when the key $K$ is opened?

Vedclass Test Series

Exam Paper Generator

Online Exam Module