Two resistances {R_1} and {R_2} are connected | vedclass

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

Question

(A) When battery ${E_2}$ is short-circuited,the terminals across which ${R_2}$ is connected are joined by a wire of zero resistance.This effectively b

Vedclass · Accepted Answer

${E_1}/{R_1}$

Vedclass · Answer

(A) When battery ${E_2}$ is short-circuited,the terminals across which ${R_2}$ is connected are joined by a wire of zero resistance.This effectively bypasses the resistor ${R_2}$,meaning no current flows through ${R_2}$ because the path of least resistance is the short-circuit wire.Consequently,the circuit simplifies to a single loop containing the battery ${E_1}$ and the resistor ${R_1}$.According to Ohm's law,the current $I$ flowing through the resistor ${R_1}$ is given by $I = {E_1}/{R_1}$.

Two resistances ${R_1}$ and ${R_2}$ are connected as shown in the figure to two batteries of $e.m.f.$ ${E_1}$ and ${E_2}$. If ${E_2}$ is short-circuited,the current through ${R_1}$ is

Similar Questions

$A$ constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if

$A$ circuit containing five resistors is connected to a battery with a $12 \,V$ emf as shown in the figure. The potential difference across the $4 \,\Omega$ resistor is ........... $V$.

In the figure shown,the capacity of the capacitor $C$ is $2\,\mu F$. The current in the $2\,\Omega$ resistor is ............... $A$.

The readings of ammeters $A_1$ and $A_2$ will be respectively:

If the galvanometer reading is zero,then the resistance $X = \dots \Omega$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module