A resistor $R$ is connected to a cell of $e m f$ $E$ and internal resistance $r$ . Potential difference across resistor $R$ is found to be $V$ . State the relation between $E, V, R$ and $r$ .

Open in App

Solution

we have formulae,

$E = I (R + r)$

$V = I R$

$∴ \frac{E}{V} = \frac{I (R + r)}{I R} = \frac{R + r}{R}$

$\frac{E}{V} - 1 = \frac{r}{R}$

$\Rightarrow r = R (\frac{E}{V} - 1)$

Suggest Corrections

Similar questions

The potential difference across a cell of emf E and internal resistance r is

Q. A cell of emf 'E' is connected across a resistance R. The potential difference between the terminals of the cell is found to be V. The internal resistance of the cell must be

Q. A cell of emf E is connected across a resistance R. The potential difference between the terminals of the cell is found to be V. The internal resistance of the cellis given as

Q. A cell of emf

E

and internal resistance

r

is connected across a resistance

r

. The potential difference between the terminals of the cell must be

Two resistors of resistance 2R and 3R are connected in series with a battery, which has an emf of E and an internal resistance R. What is the potential difference across the resistor of resistance 2R?

Join BYJU'S Learning Program

A resistor R is connected to a cell of emf E and internal resistance r. Potential difference across resistor R is found to be V. State the relation between E,V,R and r.

we have formulae,E=I(R+r)V=IR∴EV=I(R+r)IR=R+rREV−1=rR⇒r=R(EV−1)

A resistor $R$ is connected to a cell of $e m f$ $E$ and internal resistance $r$ . Potential difference across resistor $R$ is found to be $V$ . State the relation between $E, V, R$ and $r$ .

we have formulae,

$E = I (R + r)$

$V = I R$

$∴ \frac{E}{V} = \frac{I (R + r)}{I R} = \frac{R + r}{R}$

$\frac{E}{V} - 1 = \frac{r}{R}$

$\Rightarrow r = R (\frac{E}{V} - 1)$