Practice question
Question
Two cells of emf \( 5 \, \text{V} \) and \( 7 \, \text{V} \) with internal resistances \( 2 \, \Omega
\) and \( 4 \, \Omega \) are connected in series with a \( 6 \, \Omega \) resistor. What is the current
through the circuit?
Explanation
**Kirchhoff's loop rule** (energy conservation) states algebraic sum of potential differences around closed loop zero, Σ ε - Σ I R =0, ensures total voltage drop across series resistors equals source emf. Junction rule (charge conservation) Σ I_in = Σ I_out ensures current continuity. Equivalent emf: εₑq = 5 + 7 = 12 V . Total resistance: Rtₒtₐl = 2 + 4 + 6 = 12 Ω . Current: I = (εₑq/Rtₒtₐl) = (12/12) = 1 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 series/parallel, V = ε - I r and P
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