A cell of emf E and internal resistance r is connected in series with an external resistance nr . Then, the ratio of the terminal potential difference to emf is (n/n + α ) . Write the value of α .

Question

A cell of emf E and internal resistance r is connected in series with an external resistance nr . Then, the ratio of the terminal potential difference to emf is (n/n + α ) . Write the value of α . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

For the given circuit, i=(E/(n + 1) r) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-ccvuj7mhoas1nnka.jpg" /> Potential drop across battery, VT=i(n r)= (E/(n + 1) r)× nr Hece, (VT/E)=(n/n + 1)

Q. A cell of emf $E$ and internal resistance $r$ is connected in series with an external resistance $n r$ . Then, the ratio of the terminal potential difference to emf is $\frac{n}{n + α}$ . Write the value of $α$ .

For the given circuit, $i = \frac{E}{( n + 1 ) r}$

Potential drop across battery, $V_{T} = i (n r) = \frac{E}{( n + 1 ) r} \times n r$
Hece, $\frac{V _{T}}{E} = \frac{n}{n + 1}$

Q. A cell of emf E and internal resistance r is connected in series with an external resistance nr . Then, the ratio of the terminal potential difference to emf is n+αn​ . Write the value of α .

For the given circuit, i=(n+1)rE​ Potential drop across battery, VT​=i(nr)=(n+1)rE​×nr Hece, EVT​​=n+1n​

Q. A cell of emf $E$ and internal resistance $r$ is connected in series with an external resistance $n r$ . Then, the ratio of the terminal potential difference to emf is $\frac{n}{n + α}$ . Write the value of $α$ .

For the given circuit, $i = \frac{E}{( n + 1 ) r}$

Potential drop across battery, $V_{T} = i (n r) = \frac{E}{( n + 1 ) r} \times n r$
Hece, $\frac{V _{T}}{E} = \frac{n}{n + 1}$