The reading of the ammeter in the following f | vedclass

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

Question

(C) From the circuit diagram, the resistors $2 \Omega$, $3 \Omega$, and $6 \Omega$ are connected in parallel.Let their equivalent resistance be $R_p$.

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(C) From the circuit diagram, the resistors $2 \Omega$, $3 \Omega$, and $6 \Omega$ are connected in parallel.Let their equivalent resistance be $R_p$. Then, $\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1 \Omega^{-1}$.Thus, $R_p = 1 \Omega$.This parallel combination is in series with the $4 \Omega$ resistor and the $2 	ext{ V}$ battery.Therefore, the total equivalent resistance of the circuit is $R_{	ext{eq}} = R_p + 4 \Omega = 1 \Omega + 4 \Omega = 5 \Omega$.The current $I$ measured by the ammeter is given by Ohm's law: $I = \frac{V}{R_{	ext{eq}}} = \frac{2 	ext{ V}}{5 \Omega} = 0.4 	ext{ A}$.

The reading of the ammeter in the following figure will be: (in $\text{A}$)

Similar Questions

Consider a circuit where a cell of emf $E_0$ and internal resistance $r$ is connected across the terminals $A$ and $B$ as shown in the figure. The value of $R$ for which the power generated in the circuit is maximum is given by

The three resistances $A, B$ and $C$ have values $3R, 6R$ and $R$ respectively. When some potential difference is applied across the network,the thermal powers dissipated by $A, B$ and $C$ are in the ratio

In the given circuit,the potential of the point $E$ is

The figure shows a network of resistors and a battery. If $1\,A$ current flows through the branch $CF$,then find the $emf$ $E$ of the battery. (in $,V$)

Find the current in the three resistors as shown in the following figure.

Vedclass Test Series

Exam Paper Generator

Online Exam Module