In the given circuit,the internal resistance | vedclass

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

Question

(D) Case $1$: When the key $(K)$ is open,the circuit consists of the cell $(12 \ V)$ and one resistor of $40 \ \Omega$ in series with the ammeter. The

Vedclass · Accepted Answer

$1: 2$

Vedclass · Answer

(D) Case $1$: When the key $(K)$ is open,the circuit consists of the cell $(12 \ V)$ and one resistor of $40 \ \Omega$ in series with the ammeter. The current $i_1$ is given by Ohm's law: $i_1 = V / R = 12 / 40 = 0.3 \ A$.Case $2$: When the key $(K)$ is closed,the two $40 \ \Omega$ resistors are connected in parallel. The equivalent resistance $R_{eq}$ is: $1 / R_{eq} = 1 / 40 + 1 / 40 = 2 / 40 = 1 / 20$,so $R_{eq} = 20 \ \Omega$.The current $i_2$ is: $i_2 = V / R_{eq} = 12 / 20 = 0.6 \ A$.Therefore,the ratio $i_1: i_2 = 0.3: 0.6 = 1: 2$.

In the given circuit,the internal resistance of the cell is zero. If $i_1$ and $i_2$ are the readings of the ammeter when the key $(K)$ is opened and closed respectively,then $i_1: i_2=$

Similar Questions

The internal resistances of two cells shown in the figure are $0.1\,\Omega$ and $0.3\,\Omega$. If $R = 0.2\,\Omega$,find the potential difference across the cells.

If the galvanometer $G$ does not show any deflection in the circuit shown,the value of $R$ is given by $............\Omega$.

The reading of the voltmeter in the circuit shown is ............. $V$. (in $.25$)

Three resistors $1 \Omega, 2 \Omega$ and $3 \Omega$ are connected to form a triangle. Across the $3 \Omega$ resistor,a $3 \text{ V}$ battery is connected. The current through the $3 \Omega$ resistor is: (in $\text{ A}$)

All wires have the same resistance $r$. The equivalent resistance between $A$ and $B$ is $R$. Now,if the keys $K_1$ and $K_2$ are closed,what will be the new equivalent resistance?

Vedclass Test Series

Exam Paper Generator

Online Exam Module