To find the resistance of the resistor, we can use Ohm's law, which states that the voltage (V) across a resistor is equal to the current (I) through it times its resistance (R). This can be written as V = IR.

First, we need to find the total voltage in the circuit. The total voltage is the sum of the emf of the battery and the voltage drop across the internal resistance of the battery. The voltage drop across the internal resistance can be found by multiplying the current by the internal resistance (I*R_internal), which is 0.5 A * 3 Ω = 1.5 V.

The total voltage in the circuit is then 10 V (emf of the battery) - 1.5 V (voltage drop across the internal resistance) = 8.5 V.

Now we can find the resistance of the resistor. Rearranging Ohm's law to solve for R gives us R = V/I. Substituting the values we have, we get R = 8.5 V / 0.5 A = 17 Ω.

The terminal voltage of the battery when the circuit is closed is the voltage across the resistor, which is 8.5 V.

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To find the resistance of the resistor, we can use Ohm's law, which states that the voltage (V) across a resistor is equal to the current (I) through it times its resistance (R). This can be written as V = IR.

First, we need to find the total voltage in the circuit. The total voltage is the sum of the emf of the battery and the voltage drop across the internal resistance of the battery. The voltage drop across the internal resistance can be found by multiplying the current by the internal resistance (I*R_internal), which is 0.5 A * 3 Ω = 1.5 V.

The total voltage in the circuit is then 10 V (emf of the battery) - 1.5 V (voltage drop across the internal resistance) = 8.5 V.

Now we can find the resistance of the resistor. Rearranging Ohm's law to solve for R gives us R = V/I. Substituting the values we have, we get R = 8.5 V / 0.5 A = 17 Ω.

The terminal voltage of the battery when the circuit is closed is the voltage across the resistor, which is 8.5 V.

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To find the resistance of the resistor, we can use Ohm's law, which states that the voltage (V) across a resistor is equal to the current (I) through it times its resistance (R). This can be written as V = IR.

First, we need to find the total voltage in the circuit. The total voltage is the sum of the emf of the battery and the voltage drop across the internal resistance of the battery. The voltage drop across the internal resistance can be found by multiplying the current by the internal resistance (I*R_internal), which is 0.5 A * 3 Ω = 1.5 V.

The total voltage in the circuit is then 10 V (emf of the battery) - 1.5 V (voltage drop across the internal resistance) = 8.5 V.

Now we can find the resistance of the resistor. Rearranging Ohm's law to solve for R gives us R = V/I. Substituting the values we have, we get R = 8.5 V / 0.5 A = 17 Ω.

The terminal voltage of the battery when the circuit is closed is the voltage across the resistor, which is 8.5 V.

A battery of emf 10 V and internal resistance 3 W is connected to aresistor. If the current in the circuit is 0.5 A, what is the resistanceof the resistor? What is the terminal voltage of the battery when thecircuit is closed?

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