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MHT-CET Physics · Formula sheet

Current Electricity formulas

9 formulas and 15 common traps for MHT-CET Physics Current Electricity, grouped by subtopic.

Full notes with worked examples

E.m.f., Internal Resistance and Cells Together

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Terminal Voltage and Cells in Series or Parallel

Terminal voltage

V=E−Ir,I=ER+rV = E - Ir, \qquad I = \frac{E}{R + r}

Common traps

Reading e.m.f. and terminal voltage as the same

A voltmeter across a cell that is delivering current reads E − Ir, less than the e.m.f. The e.m.f. appears only when no current flows.

Kirchhoff's Current and Voltage Laws

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The Junction Law

Junction law

∑Iin=∑Iout\sum I_{\text{in}} = \sum I_{\text{out}}

The Loop Law and Potential Differences

Loop law

∑E=∑IR\sum E = \sum IR

Common traps

Swapping the conservation laws

The JUNCTION law conserves charge; the LOOP law conserves energy. 'Kirchhoff's second law' is the loop law.

Getting the sign of a cell wrong along a branch

Going from A to B, a cell entered at its + terminal is a DROP of E; entered at − it is a rise. Draw the path and mark each sign before adding.

The Wheatstone Bridge and the Metre Bridge

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The Balanced Bridge

Balance condition

PQ=RS\frac{P}{Q} = \frac{R}{S}

The Metre Bridge and its Null Point

Metre bridge

XR=l100−l\frac{X}{R} = \frac{l}{100 - l}

Common traps

Solving a balanced bridge by Kirchhoff's laws

Check the ratio first. If P/Q = R/S the middle arm carries nothing and can be deleted, whatever its resistance — the network collapses to two series pairs in parallel.

Assuming a bridge is balanced because it looks symmetric

Balance is a ratio, not a shape. Arms of 4, 4, 1 and 3 Ω are not balanced, and current flows through the galvanometer toward the lower-potential junction.

Thinking a thicker wire moves the null point

Every centimetre of the new wire has the same resistance as every other, so the ratio of the two lengths — the only thing the balance reads — is unchanged. The null point stays at l.

Measuring from the wrong end

l is measured from the end next to the gap it is paired with. 'From the centre' means |l − 50|: 40 Ω and 60 Ω balance at 40 cm, which is 10 cm left of the centre.

The Potentiometer

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Potential Gradient and the Balancing Length

Potential gradient

k=ERw+R+r⋅RwL,E′=klk = \frac{E}{R_w + R + r} \cdot \frac{R_w}{L}, \qquad E' = kl

Comparing E.m.f.s and Finding Internal Resistance

Internal resistance

r=R l0−llr = R\,\frac{l_0 - l}{l}

Common traps

Forgetting the series resistance in the gradient

When a resistance is in series with the wire, only the wire's share of the driving voltage sets k. Find I from the whole circuit first, then k = IR_w/L.

Thinking a longer wire moves the null point closer

At the same voltage a longer wire has a smaller gradient, so a cell needs MORE length to balance. The null point moves further along.

Using the sum formula for 'one cell, then opposed'

If the first length is E₁ ALONE, the ratio is l₁/(l₁ − l₂), not (l₁ + l₂)/(l₁ − l₂). Read which combination each length balances.

Taking the shunted length for the e.m.f.

With a shunt the cell delivers current, so the balance reads its terminal voltage, smaller than E. Only the unshunted length l₀ measures E.

Converting a Galvanometer into an Ammeter or a Voltmeter

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The Shunt: Galvanometer to Ammeter

Shunt

S=Ig GI−IgS = \frac{I_g\,G}{I - I_g}

The Series Resistance: Galvanometer to Voltmeter

Series resistance

R=VIg−GR = \frac{V}{I_g} - G

Common traps

Using the fraction through the shunt instead of the galvanometer

S = fG/(1 − f) with f the share through the GALVANOMETER. 5% through G is G/19; reading 5% as the shunt's share gives 19G, a resistance no ammeter could have.

Quoting the shunt as the ammeter's resistance

The ammeter is G and S in parallel, fG — slightly less than S. The question asks for the ammeter, not the shunt, when it says 'resistance of the ammeter'.

Forgetting to subtract G

V/I_g is the WHOLE voltmeter's resistance. The resistance to add is V/I_g − G.

Reading 'replaced by' as 'added to'

'A resistance of 1000 Ω is connected in series' to double the range usually REPLACES the 100 Ω: 2(G + 100) = G + 1000 gives G = 800 Ω. Where the question says 'in series with X', add it: G + X + 1500 = 2(G + X).

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