Four cells each of emf $1.9 V$ and the internal resistance $1 Ω$ are connected to an external resistance of $18 O m e g a$ as shown in the figure. Find the terminal potential difference of this combination.

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Solution

The potential difference is = 3.69V

Explanation:
We apply the equation,

$U = R I$
$E - U = r I$
$E = (R + r) I$

The emf is $E = 2 * 1.9 = 3.8 V$

The cells are in series-parallel and the internal resistances as well.

The total resistance is, $r + r = \frac{1}{2} + \frac{1}{2} + 18 = 19$

The current is, $i = \frac{3.9}{19} = 0.21 a$

So, the potential difference is, $U = 18 * 0.21 = 3.69 V$

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Four cells each of emf 1.9 V and the internal resistance 1Ω are connected to an external resistance of 18 Omega as shown in the figure. Find the terminal potential difference of this combination.

Four cells each of emf $1.9 V$ and the internal resistance $1 Ω$ are connected to an external resistance of $18 O m e g a$ as shown in the figure. Find the terminal potential difference of this combination.