A $9 V$ battery is connected in series with a resistor. The terminal voltage is found to be $8 V$ . Current through the circuit is measured as $5 A$ . What is the internal resistance of the battery?

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Solution

As we know battery supplied voltage $= E - I r$
Where $E$ is the EMF of cell , $I$ is the current passing through battery and $r$ is internal resistance of battery .
Putting all the values we get
$8 = 9 - 5 \times r$
$8 - 9 = 5 r$
$\Rightarrow r = \frac{1}{5} = 0.2 ω$

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A 9 V battery is connected in series with a resistor. The terminal voltage is found to be 8 V. Current through the circuit is measured as 5 A. What is the internal resistance of the battery?

As we know battery supplied voltage =E−IrWhere E is the EMF of cell , I is the current passing through battery and r is internal resistance of battery . Putting all the values we get8=9−5×r8−9=5r⇒r=15=0.2ω

A $9 V$ battery is connected in series with a resistor. The terminal voltage is found to be $8 V$ . Current through the circuit is measured as $5 A$ . What is the internal resistance of the battery?

As we know battery supplied voltage $= E - I r$
Where $E$ is the EMF of cell , $I$ is the current passing through battery and $r$ is internal resistance of battery .
Putting all the values we get
$8 = 9 - 5 \times r$
$8 - 9 = 5 r$
$\Rightarrow r = \frac{1}{5} = 0.2 ω$