Kirchhoff’s rules at work: the network of NCERT Example 3.6किरखोफ़ के नियमों का उपयोग: NCERT उदाहरण 3.6 का परिपथKirchhoff’s rules at work: the network of NCERT Example 3.6

NCERT Fig 3.17 · Class 12 Physics · Chapter 3 · interactive diagram

Read Chapter 3 with this diagram All Physics diagrams

Parts and what they mean

10 V cell in branch CA (शाखा CA में 10 V का सेल) — Its + terminal faces A, so going C → A through it the potential rises by 10 V.
1 Ω (in branch CA, with the 10 V cell) (1 Ω (शाखा CA में, 10 V सेल के साथ)) — Carries the full current I₁ = 2.5 A, so the drop across it is 2.5 V.
5 V cell in branch DEB (शाखा DEB में 5 V का सेल) — Its + terminal faces B: going D → E → B the potential rises by 5 V. It carries I₃ = 15/8 A upward.
4 Ω in arm AB (भुजा AB में 4 Ω) — Current I₂ = 5/8 A from A to B; drop 4 × 5/8 = 2.5 V.
4 Ω in arm AD (भुजा AD में 4 Ω) — Current I₁ − I₂ = 15/8 A from A to D; drop 7.5 V.
2 Ω in arm BC (भुजा BC में 2 Ω) — Current I₂ + I₃ = 5/2 A from B to C; drop 5 V.
2 Ω in arm CD (भुजा CD में 2 Ω) — No current: C and D turn out to be at the same potential.
Current in AB (AB में धारा) — I₂ = 5/8 A = 0.625 A, from A to B.
Current in AD (AD में धारा) — Junction rule at A: I₁ in = I₂ + (I₁ − I₂) out. Value 15/8 A = 1.875 A, from A to D.
Current in BC (BC में धारा) — At B, I₂ (from A) and I₃ (from the 5 V branch) join: 5/8 + 15/8 = 5/2 A, from B to C.
Current in CD (CD में धारा) — Junction rule at C gives I₂ + I₃ − I₁ = 5/2 − 5/2 = 0.
Current in DEB (DEB में धारा) — I₃ = 15/8 A = 1.875 A, from D through E to B (through the 5 V cell from − to +).
Current in CA (CA में धारा) — I₁ = 2.5 A, from C through the 1 Ω and the 10 V cell to A.

Original diagram drawn by Simple Learning Hub, based on NCERT Fig 3.17. Report a mistake