Science M.C.Q-[Phy-1M]

2. 25W

Explanation:

$\text{R}=\frac{\text{V}^{2}}{\text{P}}=\frac{(220)^{2}}{100}=484\Omega$

The resistance of the bulb remains constant

if the supply voltage is reduced to 110V then V $=110\text{V}$

 the power consumed by it is given by the expression $=\frac{\text{V}^{2}}{\text{R}}=\frac{(110)^{2}}{484}=25\text{W}$

3. 9.2A.

Explanation:

Resistance = $25Ω$

Voltage = 230V

$\text{Current}=\frac{\text{Voltage}}{\text{Resistance}}$

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

$\text{I}=\frac{230}{25}$

$\text{I}=9.2\text{A}$

Thus, if an electrical appliance has a resistance of $25Ω$ and when that electrical appliance is connected to a 230V supply line, the current passing through it will be 9.2A.

2. 4A

Explanation:

Since the resistors are connected in parallel,the voltage across each resistor is the same ,i.e 24V.

Therefore, current in each resistor can now simply be calculated using the relation,

$\text{I}=\frac{\text{V}}{\text{R}}\Rightarrow\text{I}=\frac{24}{6.0}=4.0\text{ A}$

3. $2\text{A}$

Explanation: $\text{R}=\rho\frac{\text{l}}{\text{A}}\ ....(\text{i})$

Also, $\text{R}=\frac{\rho(2\text{l})}{\text{A}'}\ ....(\text{ii})$

From eq. (i) and (ii)

$\frac{1}{\text{A}}=\frac{2}{\text{A}'}\text{ or A}'=2\text{A}$

1. 10²⁰

Explanation: $\text{I}=\frac{\text{Q}}{\text{t}}=\frac{\text{ne}}{\text{t}}$

$\text{n}=\frac{\text{It}}{\text{e}}=\frac{1\times16}{1.6\times10^{-19}}=10^{20}$​​​​​​​

4. $24Ω$

Explanation:

Voltage, V = 12V

Current, I = 0.5A
Resistance, $\text{R}=\frac{\text{Voltage}}{\text{Current}}$ 

$\text{R}=\frac{12}{0.5}$

$\text{R} = 24Ω$

Thus, if a car's headlight bulb working on a 12V car battery draws a current of 0.5A, the resistance of the light bulb will be $ 24Ω$