Question

please please please solve them ...... ​

Answer 1

17.

• Current, $\sf{I=0.8\ A}$

• Potential Difference, $\sf{V=20\ V}$

By Ohm's Law, resistance,

$\displaystyle\longrightarrow\sf{R=\dfrac{V}{I}}$

$\displaystyle\longrightarrow\sf{R=\dfrac{20}{0.8}\ \Omega}$

$\displaystyle\longrightarrow\sf{\underline{\underline{R=25\ \Omega}}}$

18.

• Resistance, $\sf{R=2\ k\Omega=2\times10^3\ \Omega}$

• Current, $\sf{I=10\ mA=10^{-2}\ A}$

By Ohm's Law, potential difference,

$\displaystyle\longrightarrow\sf{V=IR}$

$\displaystyle\longrightarrow\sf{V=2\times10^3\times10^{-2}\ V}$

$\displaystyle\longrightarrow\sf{\underline{\underline{V=20\ V}}}$

19.

• Current, $\sf{I=50\ mA=50\times10^{-3}\ A}$

• Voltage, $\sf{V=12\ V}$

By Ohm's Law, resistance,

$\displaystyle\longrightarrow\sf{R=\dfrac{V}{I}}$

$\displaystyle\longrightarrow\sf{R=\dfrac{12}{50\times10^{-3}}\ \Omega}$

$\displaystyle\longrightarrow\sf{\underline{\underline{R=240\ \Omega}}}$

20.

• Initial voltage, $\sf{V_1=100\ V}$

• Initial current, $\sf{I_1=5\ mA=5\times10^{-3}\ A}$

By Ohm's Law, resistance,

$\displaystyle\longrightarrow\sf{R=\dfrac{V_1}{I_1}}$

$\displaystyle\longrightarrow\sf{R=\dfrac{100}{5\times10^{-3}}\ \Omega}$

$\displaystyle\longrightarrow\sf{\underline{\underline{R=20000\ \Omega}}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{R=20\ k\Omega}}}$

• New voltage, $\sf{V_2=25\ V}$

Resistance remains unchanged. Hence new value of current flow is,

$\displaystyle\longrightarrow\sf{I_2=\dfrac{V_2}{R}}$

$\displaystyle\longrightarrow\sf{I_2=\dfrac{25}{20000}\ A}$

$\displaystyle\longrightarrow\sf{\underline{\underline{I_2=1.25\times10^{-3} A}}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{I_2=1.25\ mA}}}$

21.

• Voltage, $\sf{V=120\ V}$

• Current in the first case, $\sf{I_a=50\ mA=50\times10^{-3}\ A}$

Then by Ohm's Law, the resistance in first case is,

$\displaystyle\longrightarrow\sf{R_a=\dfrac{V}{I_a}}$

$\displaystyle\longrightarrow\sf{R_a=\dfrac{120}{50\times10^{-3}}\ \Omega}$

$\displaystyle\longrightarrow\sf{\underline{\underline{R_a=2400\ \Omega}}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{R_a=2.4\ k\Omega}}}$

• Current in the second case, $\sf{I_b=200\ \mu A=200\times10^{-6}\ A}$

Then by Ohm's Law, the resistance in second case is,

$\displaystyle\longrightarrow\sf{R_b=\dfrac{V}{I_b}}$

$\displaystyle\longrightarrow\sf{R_b=\dfrac{120}{200\times10^{-6}}\ \Omega}$

$\displaystyle\longrightarrow\sf{\underline{\underline{R_b=600000\ \Omega}}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{R_b=0.6\ M\Omega}}}$