RBSE Class 12 Physics Chapter 3 — Current Electricity: Most Repeated Questions (2013–2026)

Kirchhoff's junction rule (or Kirchhoff's first rule, or the current rule) states that the algebraic sum of the currents meeting at any junction in a circuit is zero

∑I_in =∑I_out

If current through galvanometer (I_g) is zero, then

P/Q=R/S

Proof:

In loop ABD: (Applying Kirchhoff's Voltage Law)

I₁P+I_gR_g-I₂Q=0 …(1)

In loop BCD: (Applying Kirchhoff's Voltage Law)

I_gR_g+S(I₂+I_g)-R(I₁-I_g)=0

At Balancing Condition:

I_g=0

From (1):

I₁P+(0)R_g-I₂Q=0

I₁P-I₂Q=0

I₁P=I₂Q …(A)

From (2):

0+S(I₂+0)-R(I₁-0)=0

SI₂-RI₁=0

RI₁=SI₂ …( B)

Dividing (A) and (B):

I₁P/RI₁=I₂Q/SI₂

P/R=Q/S

Or

P/Q=R/S

All solved questions for this chapter are in the full RBSE Physics PYQ book.

V=E-Ir

V₁=E₁-Ir₁; V₂=E₂-Ir₂

V_m=E_m-Ir_m

V_eq=V₁+V₂+…+V_m; V_eq=(E₁-Ir₁)+(E₂-Ir₂)+…+(E_m-Ir_m); V_eq=(E₁+E₂+E₃+…)-I(r₁+r₂+r₃+…)

V_eq=E_eq-Ir_eq

On comparing the equations:

E_eq=E₁+E₂+E₃+…

r_eq=r₁+r₂+r₃+…

All solved questions for this chapter are in the full RBSE Physics PYQ book.

Mobility

All solved questions for this chapter are in the full RBSE Physics PYQ book.

The algebraic sum of all the potential differences (voltages) around any closed loop in a circuit is zero.”

All solved questions for this chapter are in the full RBSE Physics PYQ book.

Kirchoff's Junction law:

Total current entering the junction always equals to total current Leaving the Junction

Current Split (A): Current (I) enters at A and splits equally into three branches along the edges: A to B, A to D, and A to E

Current in each of these three branches is I/3

Symmetry: Due to symmetry, the potential at the three points adjacent to A(B,D,E) is the same. Similarly, the potential at the three points adjacent to G(C,F,H) is the same

Current Recombination (G): Current recombines equally from three branches at G

The middle layer (B, D, E to C, F, H) connects the I/3 branches to the final I/3 branches

Equivalent Resistance Formula: The equivalent resistance (R_eq ) between diagonally opposite corners of a cube with all edges having resistance R is:

R_eq=5/6R

Given

R=2Ω

R_eq=5/6×2Ω

R_eq=10/6Ω

R_eq ≈1.667Ω

All solved questions for this chapter are in the full RBSE Physics PYQ book.