मराठी
CISCEISC (Science) ISC Class 12
प्रश्नपत्रिका३७४
पाठ्यपुस्तक उत्तरे१४१५०
महत्वाचे प्रश्न आणि उत्तरे६१५२
संकल्पना स्पष्टीकरण आणि व्हिडिओ८५६
टाइम टेबल१६
अभ्यासक्रम

Relation between E, V, and r

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Estimated time: 7 minutes
CISCE: Class 12

Discharging a Cell (Cell Supplying Current)

When a cell of emf E and internal resistance r is connected to an external resistance R, current I flows through the circuit.

Derivation:

E = V + Ir
V = E − Ir
E = I(R + r)
I = \[\frac {E}{R+r}\]
r = R(\[\frac {E}{V}\] − 1)
Analogy: Internal resistance is like friction inside a water pump — some pressure (energy) is lost pushing water through the pump itself before it reaches the pipe (external circuit), so the pressure delivered outside (V) is always less than the pump's total capacity (E).
CISCE: Class 12

Charging a Cell (Current Forced Into the Cell)

When an external source forces current into the cell (opposite direction), the relation changes:

V = E + Ir
Common Misconception: Students often assume V is always less than E. This is true only during discharge. During charging, V is greater than E because energy must be supplied to overcome both the emf and internal resistance
CISCE: Class 12

Special Cases

Condition Formula Physical Meaning
Open circuit (I = 0) V = E No current flows, so no voltage drop across r; terminal voltage equals emf
Negligible internal resistance (r ≈ 0) V = E Ideal cell behaviour; rarely true in practice
R → ∞ (very large external resistance) V → E Current becomes negligible, terminal voltage approaches emf
Charging condition V = E + Ir Terminal voltage exceeds emf while charging
Short circuit (R = 0) I = E/r Maximum possible current; terminal voltage drops to zero
CISCE: Class 12

Power Delivered and Maximum Power Transfer

Power delivered to the external resistance:

P = I2R = \[\frac {E^2R}{(R+r)^2}\]

Maximum power is transferred when R = r, giving:

Pmax = \[\frac {E^2}{4r}\]
CISCE: Class 12

Example 1

Given: Cell emf E = 1.5 V, internal resistance r = 0.10 Ω, current I = 2.0 A

Find: Chemical energy consumed per second, energy dissipated internally, energy delivered to external resistor, and output power

Solution:

  • Total power generated: Ptotal = EI = 1.5 × 2.0 = 3.0 W
  • Power dissipated internally: Pinternal = I2r = (2.0)2 × 0.10 = 0.4 W
  • Power delivered to external circuit (output power): Poutput = Ptotal − Pinternal = 3.0 − 0.4 = 2.6 W

Answer: Output power = 2.6 W; internal dissipation = 0.4 W

CISCE: Class 12

Example 2

Given: Battery emf E = 10 V, internal resistance r = 3 Ω, current I = 0.5 A

Find: External resistance R and terminal voltage V

Solution:

  • Using E = I(R + r): 10 = 0.5(R + 3) → R + 3 = 20 → R = 17Ω
  • Terminal voltage: V = E − Ir = 10 − (0.5 × 3) = 8.5 V

Answer: R = 17 Ω, V = 8.5 V

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