Consider the circuit given here with the foll | vedclass

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

Question

(A) The current $i$ in the circuit is given by $i = \frac{E}{R + r} = \frac{12}{4 + 2} = \frac{12}{6} = 2\, A$.$1$. Rate of energy loss in the source

Vedclass · Accepted Answer

Rate of energy loss in the source is = $8\, W$

Vedclass · Answer

(A) The current $i$ in the circuit is given by $i = \frac{E}{R + r} = \frac{12}{4 + 2} = \frac{12}{6} = 2\, A$.$1$. Rate of energy loss in the source (internal resistance) is $P_{loss} = i^2 r = (2)^2 	imes 2 = 4 	imes 2 = 8\, W$.$2$. Rate of energy conversion in the source (total power generated) is $P_{gen} = E 	imes i = 12 	imes 2 = 24\, W$.$3$. Power output across the external resistor $R$ is $P_{out} = i^2 R = (2)^2 	imes 4 = 4 	imes 4 = 16\, W$.$4$. Potential drop across $R$ is $V = iR = 2 	imes 4 = 8\, V$.Comparing these with the options,statement $(A)$ is correct.

Consider the circuit given here with the following parameters: $E.M.F.$ of the cell = $12\, V$,internal resistance of the cell = $2\,\Omega$,and external resistance $R = 4\,\Omega$. Which one of the following statements is true?

Similar Questions

In the circuit shown, if the current through the resistor $R$ is $\frac{1}{5} \, A$, the value of $R$ is (in $\Omega$)

In the circuit shown,the reading of the ammeter (ideal) is the same with both switches open as with both closed. Find the value of resistance $R$ in $\Omega$.

The e.m.f. of a cell is $E \text{ volts}$ and its internal resistance is $r \text{ ohm}$. The resistance in the external circuit is also $r \text{ ohm}$. The potential difference across the cell will be

The reading of the ammeter in the adjoining figure will be .............. $A$.

If the galvanometer $G$ does not show any deflection in the circuit shown,the value of $R$ is given by $............\Omega$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module