Electromotive force (e.m.f.), E = energy supplied to each coulomb of charge that flows from the source.
E= =
Since its unit is V, from V =
V1 = V G I1 R1 = IG RG
E=
E = I (R + r) =1+
E= 2.
=
11. Potentiometer (When G = 0) Diagram
Itotal = I1 = IG =
=
=
Formula R=
Potential difference between two points V=
= =constant =
3.
Maximum current Imaximum = , r = internal resistance Due to presence of r , V < E, hence
4. 5.
Electrical power, P Series – parallel Type Maximum Series Voltage Parallel Current 6. Maximum power 2
P=I R=( 7.
V = E – Ir V = IR = =
P = IV = I2R =
= constant
Hence,
=
V = IR = = 2
) R , when R = r
= constant
Thus,
Kirchhoff’s first law: The algebraic sum of the currents at a junction of a circuit is zero. [ Principle of conservation of charge ]
=
Hence, =
= ,
∑ = 0 or Iin = Iout
=
Kirchhoff’s second law: In a closed loop, the algebraic sum of the e.m.f.s is equal to the products of current and resistance. [Principle of conservation of energy ] *Refer to Kirchhoff’s Law question analysis
=
, V1 = E
=
=1+
8.
9.
Potential and current divider Poetntial divider
Itotal = I1 = I2 =
=
=
10. Shunt and multiplier Shunt Resistor of lower resistance than that of galvanometer and is connected parallel to galvanometer
∑
=∑
Current divider
Vtotal = V1 = V2 Itotal R total = I1 R1 = I2 R2 Multiplier Resistor of high resistance connected in series with the galvanometer.
2 = new length when S1 is closed
=
=
= 12. Wheatstone bridge & Metre bridge If R and S are replaced by wires. Then: PQRS are resistance
