8. When a 12 V battery is connected across an unknown resistor, there

1.	8. When a 12 V battery is connected across an unknown resistor, there is a currentof 2.5 mA in the circuit. Find the value of the resistance of the resistor.9. A battery of 9 V is connected in series with resistors of 0.2 12, 0.3 12, 0.412.0.5 12and 12 22. respectively. How much current would flow through the 12 12 resistor?
Answer» Answer: Explanation: 8. _________________________________________ Given VOLTAGE = 12V Given current = $\\\tt 2.5 x 10^{-3} A$ Using ohm's law: $\\\tt V= IR \\$ Where V= voltage, I = current and R = resistance, $\\\tt R = \dfrac{V}{I} =\dfrac{12}{2.5x10^{-3}} \Omega = 4.8k\Omega$ _________________________________________ 9. _________________________________________ Given voltage = 9V Resistances used: 0.2, 12, 0.3, 12, 0.412, 0.5, 12,12.22 (in ohms i ASSUME) Net resistance used = (0.2+ 12+ 0.3+ 12+0.412+0.5+12+12.22) $\Omega$ = 49.632 Since they are in series there will be same current flowing through each resistor but the potential difference across each would vary, Net current in the circuit = $\\\tt \dfrac{V}{R} = \dfrac{9}{49.632} A = 0.18A$ Current THOUGH both RESISTORS will be 0.18A. _________________________________________

8. When a 12 V battery is connected across an unknown resistor, there is a currentof 2.5 mA in the circuit. Find the value of the resistance of the resistor.9. A battery of 9 V is connected in series with resistors of 0.2 12, 0.3 12, 0.412.0.5 12and 12 22. respectively. How much current would flow through the 12 12 resistor?

Answer»

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Given VOLTAGE = 12V

Given current = $\\\tt 2.5 x 10^{-3} A$

Using ohm's law:

$\\\tt V= IR \\$

Where V= voltage, I = current and R = resistance,

$\\\tt R = \dfrac{V}{I} =\dfrac{12}{2.5x10^{-3}} \Omega = 4.8k\Omega$

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Given voltage = 9V

Resistances used: 0.2, 12, 0.3, 12, 0.412, 0.5, 12,12.22 (in ohms i ASSUME)

Net resistance used = (0.2+ 12+ 0.3+ 12+0.412+0.5+12+12.22) $\Omega$ = 49.632

Since they are in series there will be same current flowing through each resistor but the potential difference across each would vary,

Net current in the circuit = $\\\tt \dfrac{V}{R} = \dfrac{9}{49.632} A = 0.18A$

Current THOUGH both RESISTORS will be 0.18A.

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