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Kirchhoff's second law is considered one of the most asked concept.
81 Questions around this concept.
In the given circuit, with a steady current, the potential drop across the capacitor must be
The Kirchhoff’s first law $\left(\sum i=0\right)$ and second law $\left(\sum i R=\sum E\right)$ where the symbols have their usual meanings, are respectively based on
For the network shown below, the value of $V_B-V_A$ is __________$V$.
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The given potentiometer has its wire of resistance $10 \Omega$. When the sliding contact is in the middle of the potentiometer wire, the potential drop across the $2 \Omega$ resistor is:
The magnitude and direction of the current in the circuit shown will be:
The value of R is
Current flowing through the circuit is
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If the current through the circuit is 2A. Find E.
The electric potential at O is
Kirchoff's second law
The algebraic sum of all the potential across a closed loop is zero. This law is also known as Kirchhoff's Voltage law (KVL)
Change in Potential in traversing a resistance is -iR
Change in Potential in the opposite direction is iR
Traversing an e.m.f source from negative to positive terminal is -E
While in the opposite direction is E
Change in Potential in traversing a capacitor from negative to positive $\frac{q}{C}$
While in the opposite direction is $-\frac{q}{C}$
In closed-loop
$-i_1 R_1+i_2 R_2-E_1-i_3 R_3+E_2+E_3-i_4 R_4=0$
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