Question

A 100 V battery is connected across a resistor and causes 5 mA current to flow. Determine the resistance of the resistor. If the voltage is now reduced to 25 V, what will be the new value of the current flowing?

Answer 1

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Answer 2

Answer:

The resistance of the resistor, R = $20,000\Omega$.

The new current flow after the voltage is reduced, I₂ = $1.25mA$.

Explanation:

Given,

The initial voltage of the battery, V₁ = $100V$

Initially, the current flow, I₁ = $5mA$ = $0.005A$

The voltage after reduction, V₂ = $25V$

The resistance of the resistor, R =?

The new current flow after the voltage is reduced, I₂ =?

As we know,

• $V = IR$
• R = $\frac{V_{1} }{I_{1} }$

After putting the values in the equation, we get:

• Resistance, R = $\frac{100}{0.005}$ = $20,000\Omega$

Now, current flow after the voltage is reduced to $25V$;

• $I_{2} = \frac{V_{2} }{R}$
• $I_{2} = \frac{25 }{20000}$ = $0.00125A$ or $1.25mA$.

Hence, the new current flow after the voltage is reduced, I₂ = $1.25mA$.