In the given figure, an ammeter reads 5 , A a | vedclass

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(B) Let the total current measured by the ammeter be $I = 5 \, A$. Let $I_1$ be the current flowing through the resistor $R$ and $I_2$ be the current

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greater than $8 \, \Omega$

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(B) Let the total current measured by the ammeter be $I = 5 \, A$. Let $I_1$ be the current flowing through the resistor $R$ and $I_2$ be the current flowing through the voltmeter. According to Kirchhoff's current law, $I = I_1 + I_2$. Therefore, $I_1 = I - I_2 = 5 - I_2$. Since the voltmeter has a very high resistance, a small current $I_2$ flows through it, so $I_2 > 0$. This implies $I_1 The voltage across the resistor $R$ is $V = 40 \, V$. Using Ohm's law, $I_1 = V / R = 40 / R$. Substituting this into the inequality $I_1 Solving for $R$, we get $R > 40 / 5$, which means $R > 8 \, \Omega$.

In the given figure, an ammeter reads $5 \, A$ and a voltmeter reads $40 \, V$. The actual value of resistance $R$ is

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