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(B) The resistance $R$ is given by the ratio of the potential difference across the resistor to the current flowing through it.$R = \frac{V_R}{I_R}$In

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Greater than $5 \ \Omega$

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(B) The resistance $R$ is given by the ratio of the potential difference across the resistor to the current flowing through it.$R = \frac{V_R}{I_R}$In the given circuit,the voltmeter is connected in parallel with the resistor $R$. Therefore,the potential difference across the resistor is $20 \ V$.However,the ammeter is connected in series with the parallel combination of the voltmeter and the resistor. The total current $I = 4 \ A$ splits into two parts: a current $I_V$ flowing through the voltmeter and a current $I_R$ flowing through the resistor.Thus,$I_R = I - I_V = 4 \ A - I_V$.Since $I_V > 0$,the current flowing through the resistor $I_R$ must be less than $4 \ A$.Therefore,$R = \frac{20 \ V}{I_R} > \frac{20 \ V}{4 \ A} = 5 \ \Omega$.Hence,the resistance $R$ is greater than $5 \ \Omega$.

$A$ student connects a voltmeter,an ammeter,and a resistor as shown in the circuit. If the voltmeter reading is $20 \ V$ and the ammeter reading is $4 \ A$,then the resistance $R$ will be:

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