Question

For Physics Newton's

Question 1
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when a 12v battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit find the value of resistance of the resistor .
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Question 2
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Several electric bulbs designed to be used on a 220 volt Electric Supply line are dated 10 watt each .how many bulb can be connected in parallel with each other across the two wires of 220 volt line with the maximum allowable current is 5 ampere .​

Answer 1

✯ Answer 1 ✯

In this problem, the current is given in milliampere (mA),so the curent is to be converted into the units of ampere (A) by dividing the given milliamperes value by 100.

Here ,

$\text{Potential\;difference,V=12V}$

$\text{Current,I = 2.5mA}$

$\sf{\implies \frac{2.5}{100} A = 0.0025 A}$

And, Resistance ,R =? (To be calculated )

From ohm's law :

$\boxed{\green{R = \frac{V}{I}}}$

So ,

$\bf{\implies R = \frac{12}{0.0025}}$

$\boxed{\boxed{Resistance,R=4800 \Omega}}$

Or

$\boxed{\green{R = 4.8 \Omega}}$

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✯ Answer 2 ✯

First we calculate the resistance (R) of the bulb by using the given power (P) and voltage (V) values.

Now,

$\boxed{\green{Power,P= \frac{{V}^{2}}{R}}}$

So ,

$\bf{\implies 10 = \frac{{220}^{2}}{R}}$

And ,

$\bf{R = \frac{220 \times 220}{10}}$

Resistance of 1 bulb ,R = 4840 ohm

We will now calculate the total resistance of the circuit by using the given (V) voltage and I (current) values.

Now,

$\bf{R =\frac{V}{I}}$

So, $\bf {R =\frac{220}{5}}$

Total Resistance = 44ohm

Thus ,the total resistance of the circuit is 44ohms .Now, all that we have to do is to find out how many 4840 ohm bulbs should be connected in parallel to obtain a total resistance of 44ohm .Suppose the number of 4840 bulbs required is x .Then :

$\bf{\frac{1}{44}={1}{4840}\times x}$

$\bf{\frac{4840}{44}}$

$\boxed {\green{x = 110}}$

Thus ,110 bulbs can be connected in parallel in this circuit .

Answer 2

✯ Answer 1 ✯

=>

X = 110.

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✯ Answer 2 ✯

=>

r = 4.8_

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HOPE IT HELP

THANKS