A generator has an e.m.f. of 440,V and an int | vedclass

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

Question

(D) The total resistance of the circuit is the sum of the internal resistance $(r)$ and the load resistance $(R)$.$R_{total} = r + R = 400\,\Omega + 4

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(D) The total resistance of the circuit is the sum of the internal resistance $(r)$ and the load resistance $(R)$.$R_{total} = r + R = 400\,\Omega + 4000\,\Omega = 4400\,\Omega$.Using Ohm's law,the current $(I)$ flowing through the circuit is given by $I = \frac{E}{R_{total}}$,where $E$ is the e.m.f.$I = \frac{440\,V}{4400\,\Omega} = 0.1\,A$.The voltage across the load $(V_{load})$ is given by $V_{load} = I 	imes R$.$V_{load} = 0.1\,A 	imes 4000\,\Omega = 400\,V$.

$A$ generator has an e.m.f. of $440\,V$ and an internal resistance of $400\,\Omega$. Its terminals are connected to a load of $4000\,\Omega$. The voltage across the load is $...........\,V$.

Similar Questions

Ratio of thermal energy released in two resistors $R$ and $3R$ connected in parallel in an electric circuit is:

Consider the circuit shown in the figure. The value of the resistance $X$ for which the thermal power generated in it is independent of small variations of its resistance is

In the given figure,each plate of the capacitor with capacitance $C$ has a charge. Find the charge on the capacitor.

What is the current $I$ shown in the given circuit?

For what value of resistance $R$ in the given circuit will the galvanometer reading be zero (in $Omega$)? The internal resistance of both batteries is negligible.

Vedclass Test Series

Exam Paper Generator

Online Exam Module