Question 471: Current Electricity

Question

In the circuit shown in the figure, $P 
eq R$. The reading of the galvanometer remains the same with switch ' $S$ ' open or closed. Then

Accepted Answer

In the circuit shown, there is no resistor connected between point B \& D .
As a result. point 'B' \& point 'D' become equipotential
$$ \Rightarrow \quad V_B=V_D $$
No current will flow through the wire even if the switch is closed.
$	herefore \quad$ Resistor R \& Galvanometer will be in series.
$$ 	herefore \quad \mathrm{I}_{\mathrm{R}}=\mathrm{I}_{\mathrm{G}} $$

Answer

$\mathrm{I}_{\mathrm{Q}}=\mathrm{I}_{\mathrm{G}}$

Answer

$\quad \mathrm{I}_{\mathrm{Q}}=\mathrm{I}_{\mathrm{R}}$

Answer

$\mathrm{I}_{\mathrm{R}}=\mathrm{I}_{\mathrm{G}}$

Answer

$\mathrm{I}_{\mathrm{P}}=\mathrm{I}_{\mathrm{G}}$

Current Electricity

In the circuit shown in the figure, P \neq R. The reading of the galvanometer remains the same with switch ' S ' open or closed. Then

Choose the correct answer: