A \( 45 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, ea
**EMF ε** is work done by non-electrostatic forces per unit charge, terminal voltage V = ε - I r, r internal resistance (Ω), I current (A). When external R = r, total resistance 2r, current I = ε/2r, power in external R is I²R = ε²/4r, total power ε²/2r, so half power dissipated externally, half internally. Equivalent resistance: Rₑq = (5/6) R = (5/6) × 5 = (25/6) ≈ 4.17 Ω . Total current: I = (V/Rₑq) = (45/(25/6)) = 45 × (6/25) = 10.8 A . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq
Ref: NCERT > Physics Book > Current Electricity > EMF, Internal Resistance and Cells Combination