268458
Theelectric current \(i\) in the circuit shown is ( \(\mathrm{E}-2011\) )
1 \(6 \mathrm{~A}\)
2 \(2 \mathrm{~A}\)
3 \(3 \mathrm{~A}\)
4 \(4 \mathrm{~A}\)
Explanation:
Applying kirchoff's first law
Current Electricity
268459
In the circuit shown in the figure, the current ' \(I\) ' is
1 \(6 \mathrm{~A}\)
2 \(2 \mathrm{~A}\)
3 \(4 \mathrm{~A}\)
4 \(7 \mathrm{~A}\)
Explanation:
Using kirchoff's law
Current Electricity
268460
Four resistors \(A, B, C\) and \(D\) form a wheatstone's bridge. The bridge is balanced when \(C=100 \Omega\). If \(A\) and \(B\) are inter changed, the bridge balances for \(C=121 \Omega\). The vlaue of \(D\) is ( E-2012)
1 \(10 \Omega\)
2 \(100 \Omega\)
3 \(110 \Omega\)
4 \(120 \Omega\)
Explanation:
Using wheatstone bridge principle
Current Electricity
268461
In the circuit shown below, the ammeter reading is zero. Then the value of the resistance \(R\) is ( E-2011)
1 \(50 \Omega\)
2 \(100 \Omega_{500 \Omega}\)
3 \(200 \Omega\)
4 \(400 \Omega\)
Explanation:
Applying kirchoff's law
Current Electricity
268504
The current\(i\) drawn from the 5 volt source will be
268458
Theelectric current \(i\) in the circuit shown is ( \(\mathrm{E}-2011\) )
1 \(6 \mathrm{~A}\)
2 \(2 \mathrm{~A}\)
3 \(3 \mathrm{~A}\)
4 \(4 \mathrm{~A}\)
Explanation:
Applying kirchoff's first law
Current Electricity
268459
In the circuit shown in the figure, the current ' \(I\) ' is
1 \(6 \mathrm{~A}\)
2 \(2 \mathrm{~A}\)
3 \(4 \mathrm{~A}\)
4 \(7 \mathrm{~A}\)
Explanation:
Using kirchoff's law
Current Electricity
268460
Four resistors \(A, B, C\) and \(D\) form a wheatstone's bridge. The bridge is balanced when \(C=100 \Omega\). If \(A\) and \(B\) are inter changed, the bridge balances for \(C=121 \Omega\). The vlaue of \(D\) is ( E-2012)
1 \(10 \Omega\)
2 \(100 \Omega\)
3 \(110 \Omega\)
4 \(120 \Omega\)
Explanation:
Using wheatstone bridge principle
Current Electricity
268461
In the circuit shown below, the ammeter reading is zero. Then the value of the resistance \(R\) is ( E-2011)
1 \(50 \Omega\)
2 \(100 \Omega_{500 \Omega}\)
3 \(200 \Omega\)
4 \(400 \Omega\)
Explanation:
Applying kirchoff's law
Current Electricity
268504
The current\(i\) drawn from the 5 volt source will be
268458
Theelectric current \(i\) in the circuit shown is ( \(\mathrm{E}-2011\) )
1 \(6 \mathrm{~A}\)
2 \(2 \mathrm{~A}\)
3 \(3 \mathrm{~A}\)
4 \(4 \mathrm{~A}\)
Explanation:
Applying kirchoff's first law
Current Electricity
268459
In the circuit shown in the figure, the current ' \(I\) ' is
1 \(6 \mathrm{~A}\)
2 \(2 \mathrm{~A}\)
3 \(4 \mathrm{~A}\)
4 \(7 \mathrm{~A}\)
Explanation:
Using kirchoff's law
Current Electricity
268460
Four resistors \(A, B, C\) and \(D\) form a wheatstone's bridge. The bridge is balanced when \(C=100 \Omega\). If \(A\) and \(B\) are inter changed, the bridge balances for \(C=121 \Omega\). The vlaue of \(D\) is ( E-2012)
1 \(10 \Omega\)
2 \(100 \Omega\)
3 \(110 \Omega\)
4 \(120 \Omega\)
Explanation:
Using wheatstone bridge principle
Current Electricity
268461
In the circuit shown below, the ammeter reading is zero. Then the value of the resistance \(R\) is ( E-2011)
1 \(50 \Omega\)
2 \(100 \Omega_{500 \Omega}\)
3 \(200 \Omega\)
4 \(400 \Omega\)
Explanation:
Applying kirchoff's law
Current Electricity
268504
The current\(i\) drawn from the 5 volt source will be
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Current Electricity
268458
Theelectric current \(i\) in the circuit shown is ( \(\mathrm{E}-2011\) )
1 \(6 \mathrm{~A}\)
2 \(2 \mathrm{~A}\)
3 \(3 \mathrm{~A}\)
4 \(4 \mathrm{~A}\)
Explanation:
Applying kirchoff's first law
Current Electricity
268459
In the circuit shown in the figure, the current ' \(I\) ' is
1 \(6 \mathrm{~A}\)
2 \(2 \mathrm{~A}\)
3 \(4 \mathrm{~A}\)
4 \(7 \mathrm{~A}\)
Explanation:
Using kirchoff's law
Current Electricity
268460
Four resistors \(A, B, C\) and \(D\) form a wheatstone's bridge. The bridge is balanced when \(C=100 \Omega\). If \(A\) and \(B\) are inter changed, the bridge balances for \(C=121 \Omega\). The vlaue of \(D\) is ( E-2012)
1 \(10 \Omega\)
2 \(100 \Omega\)
3 \(110 \Omega\)
4 \(120 \Omega\)
Explanation:
Using wheatstone bridge principle
Current Electricity
268461
In the circuit shown below, the ammeter reading is zero. Then the value of the resistance \(R\) is ( E-2011)
1 \(50 \Omega\)
2 \(100 \Omega_{500 \Omega}\)
3 \(200 \Omega\)
4 \(400 \Omega\)
Explanation:
Applying kirchoff's law
Current Electricity
268504
The current\(i\) drawn from the 5 volt source will be
268458
Theelectric current \(i\) in the circuit shown is ( \(\mathrm{E}-2011\) )
1 \(6 \mathrm{~A}\)
2 \(2 \mathrm{~A}\)
3 \(3 \mathrm{~A}\)
4 \(4 \mathrm{~A}\)
Explanation:
Applying kirchoff's first law
Current Electricity
268459
In the circuit shown in the figure, the current ' \(I\) ' is
1 \(6 \mathrm{~A}\)
2 \(2 \mathrm{~A}\)
3 \(4 \mathrm{~A}\)
4 \(7 \mathrm{~A}\)
Explanation:
Using kirchoff's law
Current Electricity
268460
Four resistors \(A, B, C\) and \(D\) form a wheatstone's bridge. The bridge is balanced when \(C=100 \Omega\). If \(A\) and \(B\) are inter changed, the bridge balances for \(C=121 \Omega\). The vlaue of \(D\) is ( E-2012)
1 \(10 \Omega\)
2 \(100 \Omega\)
3 \(110 \Omega\)
4 \(120 \Omega\)
Explanation:
Using wheatstone bridge principle
Current Electricity
268461
In the circuit shown below, the ammeter reading is zero. Then the value of the resistance \(R\) is ( E-2011)
1 \(50 \Omega\)
2 \(100 \Omega_{500 \Omega}\)
3 \(200 \Omega\)
4 \(400 \Omega\)
Explanation:
Applying kirchoff's law
Current Electricity
268504
The current\(i\) drawn from the 5 volt source will be