Question

8. When a 12 V battery is connected across an unknown resistor, there is a current
of 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 12
and 12 22. respectively. How much current would flow through the 12 12 resistor?​

Answer 1

Answer:

Explanation:

8.

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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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9.

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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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Answer 2

Answer:

Given :-

• Current = 2.5 mA
• Potential difference = 12 V

To Find :-

Resistance

Solution :-

Current = 2.5 × 10^-³

Now

Resistance = Potential difference/Current

Resistance = 12/2.5 $10^{-3}$

Resistance = 4.8 kΩ

Now

Given :-

Potential difference = 9 V

To Find :-

Current flow

Solution :-

At first the total Net resistance in the battery

0.2 Ω, 12 Ω, 0.3 Ω, 12 Ω,0.412 Ω,0.5 Ω, 12 Ω, 2.22 Ω

=> 49.632 Ω

Now

We know that

V = IR

By putting value

9 = I(49.632)

9/49.632 = I

0.18 = I