Draw a schematic diagram of an electric circuit (in the "on" position) consisting of a battery of five cells of $2\, V$ each, a $5\, \Omega$ resistor, a $8\, \Omega$ resistor, a $12\, \Omega$ resistor and a plug key, all connected in series. An ammeter is put in the circuit to measure the electric current through the resistors and a voltmeter is connected so as to measure the potential difference across the $12\, \Omega$ resistor.
Calculate the reading shown by the
$(a)$ ammeter
$(b)$ voltmeter in the given electric circuit.
Calculate the reading shown by the
$(a)$ ammeter
$(b)$ voltmeter in the given electric circuit.
$(a)$ Total resistance $R = 5\, \Omega + 8\, \Omega + 12\, \Omega = 25\, \Omega$.
Total voltage $V = 5 \times 2\, V = 10\, V$.
Using Ohm's law, the current $I = \frac{V}{R} = \frac{10\, V}{25\, \Omega} = 0.4\, A$.
Therefore, the ammeter reading is $0.4\, A$.
$(b)$ The potential difference across the $12\, \Omega$ resistor is $V' = I \times R_{12} = 0.4\, A \times 12\, \Omega = 4.8\, V$.
Therefore, the voltmeter reading is $4.8\, V$.
Total voltage $V = 5 \times 2\, V = 10\, V$.
Using Ohm's law, the current $I = \frac{V}{R} = \frac{10\, V}{25\, \Omega} = 0.4\, A$.
Therefore, the ammeter reading is $0.4\, A$.
$(b)$ The potential difference across the $12\, \Omega$ resistor is $V' = I \times R_{12} = 0.4\, A \times 12\, \Omega = 4.8\, V$.
Therefore, the voltmeter reading is $4.8\, V$.
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