The resistances of the four arms P, Q, R and | vedclass

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

Question

(A) For a balanced Wheatstone's bridge, the condition is $\frac{P}{Q} = \frac{R}{S}$.Substituting the given values: $\frac{10 \, \Omega}{30 \, \Omega}

Vedclass · Accepted Answer

$0.2$

Vedclass · Answer

(A) For a balanced Wheatstone's bridge, the condition is $\frac{P}{Q} = \frac{R}{S}$.Substituting the given values: $\frac{10 \, \Omega}{30 \, \Omega} = \frac{30 \, \Omega}{90 \, \Omega}$, which simplifies to $\frac{1}{3} = \frac{1}{3}$.Since the bridge is balanced, no current flows through the galvanometer arm. Therefore, the $50 \, \Omega$ galvanometer resistance is ineffective.The circuit simplifies to two parallel branches: $(P+Q)$ and $(R+S)$ in parallel, connected in series with the internal resistance $r = 5 \, \Omega$.The equivalent resistance of the parallel part is $R_p = \frac{(P+Q)(R+S)}{(P+Q)+(R+S)} = \frac{(10+30)(30+90)}{(10+30)+(30+90)} = \frac{40 	imes 120}{40+120} = \frac{4800}{160} = 30 \, \Omega$.The total equivalent resistance of the circuit is $R_{eq} = r + R_p = 5 \, \Omega + 30 \, \Omega = 35 \, \Omega$.The current drawn from the cell is $I = \frac{E}{R_{eq}} = \frac{7 \, V}{35 \, \Omega} = 0.2 \, A$.

Similar Questions

Find the current in the $10\,\Omega$ resistance. The circuit consists of two $5\,V$ batteries with internal resistances of $2\,\Omega$ each,connected in parallel with a $10\,\Omega$ resistor.

In the circuit diagram shown below,the magnitude and direction of the flow of current respectively would be

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

In the adjoining circuit,the battery $E_1$ has an $e.m.f.$ of $12 \, V$ and zero internal resistance,while the battery $E$ has an $e.m.f.$ of $2 \, V$. If the galvanometer $G$ reads zero,then the value of the resistance $X$ in $\Omega$ is

The figure shows a network of four resistances and three batteries. The current through the branch $CF$ is ............... $A$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module