Science Answer the questions.[Phy-2M]

1. $\text{I}=\frac{6}{1+2}=2 \ \text{A}$.

Since current flowing is same in both resistors, power used in $2\Omega$ resistor

(P₁) = I²R = (2)² × 2 = 8 W.

2. Since both $12\Omega$ and $2\Omega$ are in parallel to the 4 V source, voltage across each resistor remain same.

$\therefore$ Power used in $2\Omega$ resistor

$(\text{P}_2)=\frac{\text{V}^2}{\text{R}}=\frac{{4}^2}{2}=\frac{16}{2}=8 \ \text{W}$

Comparing between the power used in both cases $=\frac{\text{P}_1}{\text{P}_2}=\frac{8\text{W}}{8\text{W}}=1$.

1. When two 6 $\Omega$ resistances are in parallel and the third is in combination to this, the equivalent resistance will be 9 $\Omega$.

2. When two 6 $\Omega$ resistances are in series and the third is in parallel to them, then it will be 4 $\Omega$.

$\frac{1}{\text{R}}=\frac{1}{176}+\frac{1}{176}\dots\text{times}=\frac{\text{n}}{176} \ \text{or} \ \text{R}=\frac{176}{\text{n}} \ \Omega$

$\text{R}=\frac{\text{V}}{\text{I}} \ \text{or} \ \frac{176}{\text{n}}=\frac{220}{5} 44\ \Omega$

$\text{n}=\frac{176}{44}=4$, resistors should be joined in parallel.

Potential difference (V) = 5V

Current (I) = 1A

Resistance (R) $=\frac{\text{V}}{\text{I}}=5\Omega$

Thus, the resistance of given resistor is 5 ohms.

$\frac{1}{\text{R}}=\frac{1}{10}+\frac{1}{1000}=\frac{(100+1)}{1000}$

$\text{R}=\frac{1000}{101}=9.9\ \text{ohm}$

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

$\text{R}=\frac{12}{(2.5\times10^{-3})}$

we know that,Potential difference $= \frac{\text{Work done}}{\text{Charge\ moved}}$

$=\frac{250}{20}=12.5$

$=1000\times\frac{\text{J}}{\text{S}}\times60\times60\text{s}$

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