Two sources of equal emf E are connected in s | vedclass

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

Question

(D) The total $emf$ of the series combination is $E_{eq} = E + E = 2E$. The total resistance of the circuit is $R_{eq} = R + R_1 + R_2$. The current i

Vedclass · Accepted Answer

$R = R_2 - R_1$

Vedclass · Answer

(D) The total $emf$ of the series combination is $E_{eq} = E + E = 2E$. The total resistance of the circuit is $R_{eq} = R + R_1 + R_2$. The current in the circuit is given by $i = \frac{2E}{R + R_1 + R_2}$.The potential difference $V$ across a source with $emf$ $E$ and internal resistance $r$ is given by $V = E - ir$. For the source with internal resistance $R_2$,the potential difference is zero:$0 = E - i R_2$$E = i R_2$Substituting the value of $i$:$E = \left( \frac{2E}{R + R_1 + R_2} \right) R_2$$1 = \frac{2 R_2}{R + R_1 + R_2}$$R + R_1 + R_2 = 2 R_2$$R = 2 R_2 - R_2 - R_1$$R = R_2 - R_1$

Two sources of equal $emf$ $E$ are connected in series to an external resistance $R$. The internal resistances of the two sources are $R_1$ and $R_2$ $(R_2 > R_1)$. If the potential difference across the source having internal resistance $R_2$ is zero,then:

Similar Questions

In the circuit shown below,the resistances are given (in $\Omega$) and the battery is assumed ideal with an electromotive force (emf) equal to $3.0 \, V$. The resistor that dissipates the most power is

If each resistance in the figure is $9 \Omega$,then the reading of the ammeter $(A)$ is (in $A$)

The equivalent resistance of the infinite network given below is:

Resistances $R_1$ and $R_2$ are joined in parallel and a current is passed so that the amount of heat liberated is $H_1$ and $H_2$ respectively. The ratio $\frac{H_1}{H_2}$ has the value:

Vedclass Test Series

Exam Paper Generator

Online Exam Module