268359
The resultant resistance of two resistors when connected in series is \(48 \mathrm{ohm}\). The ratio of their resistances is \(3: 1\). The value of each resistance is
1 \(20 \Omega, 28 \Omega\)
2 \(32 \Omega, 16 \Omega\)
3 \(36 \Omega, 12 \Omega\)
4 \(24_{\Omega}, 24_{\Omega}\)
Explanation:
\(R_{S}=R_{1}+R_{2}, R_{P}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}\) Solving for \(R_{1} \& R_{2}\)
Current Electricity
268360
The resistance of a bulb filament is \(100 \Omega\) at a temperature of \(100^{\circ} \mathrm{C}\). If its temperature coefficient of resistance be 0.005 per \({ }^{0} \mathrm{C}\), its resistance will become \(200 \Omega\) at temperature of
268359
The resultant resistance of two resistors when connected in series is \(48 \mathrm{ohm}\). The ratio of their resistances is \(3: 1\). The value of each resistance is
1 \(20 \Omega, 28 \Omega\)
2 \(32 \Omega, 16 \Omega\)
3 \(36 \Omega, 12 \Omega\)
4 \(24_{\Omega}, 24_{\Omega}\)
Explanation:
\(R_{S}=R_{1}+R_{2}, R_{P}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}\) Solving for \(R_{1} \& R_{2}\)
Current Electricity
268360
The resistance of a bulb filament is \(100 \Omega\) at a temperature of \(100^{\circ} \mathrm{C}\). If its temperature coefficient of resistance be 0.005 per \({ }^{0} \mathrm{C}\), its resistance will become \(200 \Omega\) at temperature of
268359
The resultant resistance of two resistors when connected in series is \(48 \mathrm{ohm}\). The ratio of their resistances is \(3: 1\). The value of each resistance is
1 \(20 \Omega, 28 \Omega\)
2 \(32 \Omega, 16 \Omega\)
3 \(36 \Omega, 12 \Omega\)
4 \(24_{\Omega}, 24_{\Omega}\)
Explanation:
\(R_{S}=R_{1}+R_{2}, R_{P}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}\) Solving for \(R_{1} \& R_{2}\)
Current Electricity
268360
The resistance of a bulb filament is \(100 \Omega\) at a temperature of \(100^{\circ} \mathrm{C}\). If its temperature coefficient of resistance be 0.005 per \({ }^{0} \mathrm{C}\), its resistance will become \(200 \Omega\) at temperature of
268359
The resultant resistance of two resistors when connected in series is \(48 \mathrm{ohm}\). The ratio of their resistances is \(3: 1\). The value of each resistance is
1 \(20 \Omega, 28 \Omega\)
2 \(32 \Omega, 16 \Omega\)
3 \(36 \Omega, 12 \Omega\)
4 \(24_{\Omega}, 24_{\Omega}\)
Explanation:
\(R_{S}=R_{1}+R_{2}, R_{P}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}\) Solving for \(R_{1} \& R_{2}\)
Current Electricity
268360
The resistance of a bulb filament is \(100 \Omega\) at a temperature of \(100^{\circ} \mathrm{C}\). If its temperature coefficient of resistance be 0.005 per \({ }^{0} \mathrm{C}\), its resistance will become \(200 \Omega\) at temperature of