In a Wheatstone bridge,P = 90,Omega,Q = 110,O | vedclass

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

Question

(B) Let the potential at point $A$ be $V_A$ and at point $D$ be $V_D = 0\,V$. The total emf is $4\,V$.$1$. Calculate the potential at point $B$ $(V_B)

Vedclass · Accepted Answer

$0.2$

Vedclass · Answer

(B) Let the potential at point $A$ be $V_A$ and at point $D$ be $V_D = 0\,V$. The total emf is $4\,V$.$1$. Calculate the potential at point $B$ $(V_B)$: The resistors $P$ and $Q$ are in series across the $4\,V$ source. Using the voltage divider rule:$V_A - V_B = V_A 	imes \frac{P}{P+Q} = 4 	imes \frac{90}{90+110} = 4 	imes \frac{90}{200} = 1.8\,V$.$2$. Calculate the potential at point $C$ $(V_C)$: The resistors $R$ and $S$ are in series across the $4\,V$ source. Using the voltage divider rule:$V_A - V_C = V_A 	imes \frac{R}{R+S} = 4 	imes \frac{40}{40+60} = 4 	imes \frac{40}{100} = 1.6\,V$.$3$. Calculate the potential difference between $B$ and $C$:$V_B - V_C = (V_A - V_C) - (V_A - V_B) = 1.6\,V - 1.8\,V = -0.2\,V$.The magnitude of the potential difference is $|V_B - V_C| = 0.2\,V$.

In a Wheatstone bridge,$P = 90\,\Omega$,$Q = 110\,\Omega$,$R = 40\,\Omega$,and $S = 60\,\Omega$. $A$ cell of $4\,V$ emf is connected across the input terminals. The potential difference between the diagonal points $B$ and $C$ (where the galvanometer is connected) is ............. $V$.

Similar Questions

In the circuit given here,the potentials at points $A$,$B$,and $C$ are $70 \,V$,$0 \,V$,and $10 \,V$ respectively. Then:

For the circuit shown,with $R_1 = 1.0\,\Omega$,$R_2 = 2.0\,\Omega$,$E_1 = 2\,V$ and $E_2 = E_3 = 4\,V$,the potential difference between the points $a$ and $b$ is approximately (in $V$):

In the following circuit, the magnitude of current $I_1$ is $............A$.

Five resistances are connected as shown in the figure. The effective resistance between points $A$ and $B$ is

In the circuit shown,$E_1 = 4.0 \ V, R_1 = 2 \ \Omega, E_2 = 6.0 \ V, R_2 = 4 \ \Omega$ and $R_3 = 2 \ \Omega$. The current $I_1$ is : ................ $A$

Vedclass Test Series

Exam Paper Generator

Online Exam Module