In the given circuit, the internal resistance | vedclass

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

Question

(C) In a steady state, the capacitor acts as an open circuit, so no current flows through the branch containing the capacitor.The current $i$ flows on

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) In a steady state, the capacitor acts as an open circuit, so no current flows through the branch containing the capacitor.The current $i$ flows only through the upper branch containing the two $1\, \Omega$ resistors in series.The total resistance of the circuit is $R_{eq} = 1\, \Omega + 1\, \Omega + 0.5\, \Omega = 2.5\, \Omega$.The current in the circuit is $i = \frac{V}{R_{eq}} = \frac{2.5\, V}{2.5\, \Omega} = 1\, A$.The potential difference across the capacitor is equal to the potential difference across the parallel branch containing the $2\, \Omega$ resistor. Since no current flows through the $2\, \Omega$ resistor, the potential difference across the capacitor is equal to the terminal voltage of the battery.The terminal voltage $V_{terminal} = E - ir = 2.5\, V - (1\, A 	imes 0.5\, \Omega) = 2.5\, V - 0.5\, V = 2.0\, V$.The charge on the capacitor is $Q = C 	imes V = 5\, \mu F 	imes 2.0\, V = 10\, \mu C$.

In the given circuit, the internal resistance of the battery is $0.50\, \Omega$. What is the charge on the plates of the $5\, \mu F$ capacitor in $\mu C$?

Similar Questions

Total $emf$ produced in a thermocouple does not depend on

The following figures show different combinations of identical bulb$(s)$ connected to identical battery(ies). Which option is correct regarding the total power dissipated in the circuit?

When a current of $1\, A$ is passed through a conductor whose ends are maintained at a temperature difference of $1\, ^oC$,the amount of heat evolved or absorbed is called:

The expression for thermo $e.m.f.$ in a thermocouple is given by the relation $E = 40\theta - \frac{\theta^2}{20}$,where $\theta$ is the temperature difference of two junctions. For this,the neutral temperature will be ................... $^oC$.

The cold junction of a thermocouple is at $0^{\circ} C$. The thermo e.m.f. produced in the thermocouple is given by the equation $E = 16T - 0.04T^2$, where $T$ is the temperature of the hot junction. The temperature of inversion and the neutral temperature of the thermocouple are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module