268361
The current ' \(i\) ' in the circuit given aside is
1 \(0.1 \mathrm{~A}\)
2 \(30 \Omega\)
3 \(1.0 \mathrm{~A}\)
4 \(2.0 \mathrm{~A}\)
Explanation:
use ohm's law
Current Electricity
268362
The combined resistance of two conductors in series is \(1_{\Omega}\). If the conductance of one conductor is 1.1 siemen, the conductance of the other conductor in siemen is
268363
Four conductors of resistnace \(16 \Omega\) each are connected to form a square. The equivalent resistanceacrosstwo adjacent cornersis(in ohm)
1 6
2 18
3 12
4 16
Explanation:
Concept Based
Current Electricity
268364
When two resistances are connected in parallel then the equivalent resistance is \(6 / 5 \Omega\). W hen one of the resistance is removed then the effective resistance is \(2 \Omega\). The resistance of the wire removed will be
1 3 ohm
2 2 ohm
3 \(\frac{3}{5} \mathrm{ohm}\)
4 \(\frac{6}{5} \mathrm{ohm}\)
Explanation:
\(\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{6}{5}\). If \(R_{2}\) is removed \(R_{1}=2 \Omega\) \(\frac{2 R_{2}}{2+R_{2}}=\frac{6}{5} \Rightarrow 5 R_{2}=6+3 R_{2} \Rightarrow R_{2}=3 \Omega\)
268361
The current ' \(i\) ' in the circuit given aside is
1 \(0.1 \mathrm{~A}\)
2 \(30 \Omega\)
3 \(1.0 \mathrm{~A}\)
4 \(2.0 \mathrm{~A}\)
Explanation:
use ohm's law
Current Electricity
268362
The combined resistance of two conductors in series is \(1_{\Omega}\). If the conductance of one conductor is 1.1 siemen, the conductance of the other conductor in siemen is
268363
Four conductors of resistnace \(16 \Omega\) each are connected to form a square. The equivalent resistanceacrosstwo adjacent cornersis(in ohm)
1 6
2 18
3 12
4 16
Explanation:
Concept Based
Current Electricity
268364
When two resistances are connected in parallel then the equivalent resistance is \(6 / 5 \Omega\). W hen one of the resistance is removed then the effective resistance is \(2 \Omega\). The resistance of the wire removed will be
1 3 ohm
2 2 ohm
3 \(\frac{3}{5} \mathrm{ohm}\)
4 \(\frac{6}{5} \mathrm{ohm}\)
Explanation:
\(\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{6}{5}\). If \(R_{2}\) is removed \(R_{1}=2 \Omega\) \(\frac{2 R_{2}}{2+R_{2}}=\frac{6}{5} \Rightarrow 5 R_{2}=6+3 R_{2} \Rightarrow R_{2}=3 \Omega\)
268361
The current ' \(i\) ' in the circuit given aside is
1 \(0.1 \mathrm{~A}\)
2 \(30 \Omega\)
3 \(1.0 \mathrm{~A}\)
4 \(2.0 \mathrm{~A}\)
Explanation:
use ohm's law
Current Electricity
268362
The combined resistance of two conductors in series is \(1_{\Omega}\). If the conductance of one conductor is 1.1 siemen, the conductance of the other conductor in siemen is
268363
Four conductors of resistnace \(16 \Omega\) each are connected to form a square. The equivalent resistanceacrosstwo adjacent cornersis(in ohm)
1 6
2 18
3 12
4 16
Explanation:
Concept Based
Current Electricity
268364
When two resistances are connected in parallel then the equivalent resistance is \(6 / 5 \Omega\). W hen one of the resistance is removed then the effective resistance is \(2 \Omega\). The resistance of the wire removed will be
1 3 ohm
2 2 ohm
3 \(\frac{3}{5} \mathrm{ohm}\)
4 \(\frac{6}{5} \mathrm{ohm}\)
Explanation:
\(\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{6}{5}\). If \(R_{2}\) is removed \(R_{1}=2 \Omega\) \(\frac{2 R_{2}}{2+R_{2}}=\frac{6}{5} \Rightarrow 5 R_{2}=6+3 R_{2} \Rightarrow R_{2}=3 \Omega\)
268361
The current ' \(i\) ' in the circuit given aside is
1 \(0.1 \mathrm{~A}\)
2 \(30 \Omega\)
3 \(1.0 \mathrm{~A}\)
4 \(2.0 \mathrm{~A}\)
Explanation:
use ohm's law
Current Electricity
268362
The combined resistance of two conductors in series is \(1_{\Omega}\). If the conductance of one conductor is 1.1 siemen, the conductance of the other conductor in siemen is
268363
Four conductors of resistnace \(16 \Omega\) each are connected to form a square. The equivalent resistanceacrosstwo adjacent cornersis(in ohm)
1 6
2 18
3 12
4 16
Explanation:
Concept Based
Current Electricity
268364
When two resistances are connected in parallel then the equivalent resistance is \(6 / 5 \Omega\). W hen one of the resistance is removed then the effective resistance is \(2 \Omega\). The resistance of the wire removed will be
1 3 ohm
2 2 ohm
3 \(\frac{3}{5} \mathrm{ohm}\)
4 \(\frac{6}{5} \mathrm{ohm}\)
Explanation:
\(\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{6}{5}\). If \(R_{2}\) is removed \(R_{1}=2 \Omega\) \(\frac{2 R_{2}}{2+R_{2}}=\frac{6}{5} \Rightarrow 5 R_{2}=6+3 R_{2} \Rightarrow R_{2}=3 \Omega\)
