series and parallel
class 12 physics
Answers
Answer:
Combination of Resistors
Resistors can be combined in 2 ways:-
Resistors in Series: A series circuit is a circuit in which resistors are arranged in a single chain, resulting in common current flowing through them.
Circuit Diagram
‘N’ number of resistors can be joined together.
As all the resistors are connected to each other as a result same amount of current flows through each resistor.
But the Potential difference will be different in each resistor.
Consider current flowing through all the resistors =I,Resistance across first resistor=R1.
Potential difference across resistor R1is V1, V1=IR1(By ohm’s Law)
Similarly V2=IR2, V3= IR3 and so on.
Therefore Total Voltage V=V1+V2+V3+….. +Vn
IRequivalent =IR1+IR2+IR3+…. +IRn where Requivalent is the equivalent resistance of the circuit.
=>Requivalent =(R1+ R2+ R3+….+Rn)
Therefore if the resistances are connected in series then the total equivalent resistance of the circuit is equal to the sum of all the resistors in the circuit.
Resistors in Parallel:- A parallel circuit is a circuit in which the resistors are arranged with their heads connected together, and their tails connected together.
The potential difference(V)across eachresistoris same.
The amount of current flowing is different. This means I= I1+I2+I3+…+In
=>From Ohm’s law - (V/Requivalent) = (V/R1)+(V/R2)+(V/R3)+…..+(V/Rn)
=>1/ Requivalent =(1/ R1+1/ R2+1/R3+….+1/Rn)
Therefore if the resistances are connected in parallel then the total equivalent resistance of the circuit is equal to the sum of the reciprocal of all the resistors connected in the circuit.
Circuit diagram
Problem:- A network of resistors is connected to a 16 V batterywith internal resistance of 1Ω, as shown in Fig. (a) Computethe equivalent resistance of the network. (b) Obtain the current ineach resistor. (c) Obtain the voltage drops VAB, VBC and VCD.
Answer:- (a) The network is a simple series and parallel combination ofresistors. First the two 4Ω resistors in parallel are equivalent to aresistor = [(4 × 4)/(4 + 4)] Ω = 2 Ω.
In the same way, the 12 Ω and 6 Ω resistors in parallel areequivalent to a resistor of
[(12 × 6)/(12 + 6)] Ω = 4 Ω.
The equivalent resistance R of the network is obtained bycombining these resistors (2 Ω and 4 Ω) with 1 Ω in series,that is,
R = 2 Ω + 4 Ω + 1 Ω = 7 Ω.
(b) The total current I in the circuit is
I=E/(R+r) =16V/(7+1) Ω =2A.
Consider the resistors between A and B. If I1 is the current in oneof the 4 Ω resistors and I2 the current in the other,
I1 × 4 = I2 × 4
that is, I1 = I2, which is otherwise obvious from the symmetry ofthe two arms.
But I1 + I2 = I = 2 A. Thus,
I1 = I2 = 1 A
that is, current in each 4 Ω resistor is 1 A. Current in 1 Ω resistorbetween B and C would be 2 A.
Now, consider the resistances between C and D. If I3 is the currentin the 12 Ω resistor,
and I4 in the 6 Ω resistor,
I3 × 12 = I4 × 6, i.e., I4 = 2I3
But, I3 + I4 = I = 2 A
Thus, I3 = (2/3) A,I4= (4/3)A
that is, the current in the 12 Ω resistor is (2/3) A, while the currentin the 6 Ω resistor is
(4/3) A.
(c) The voltage drop across AB is
VAB = I1 × 4 = 1 A × 4 Ω = 4 V,
This can also be obtained by multiplying the total current betweenA and B by the equivalent resistance between A and B, that is,
VAB = 2 A × 2 Ω = 4 V
The voltage drop across BC is
VBC = 2 A × 1 Ω = 2 V
Finally, the voltage drop across CD is
VCD = 12 Ω × I3 = 12 Ω × (2/3) A=8V.
This can alternately be obtained by multiplying total currentbetween C and D by the equivalent resistance between C and D,that is,
VCD = 2 A × 4 Ω = 8 V
Note that the total voltage drop across AD is 4 V + 2 V + 8 V = 14 V.
Thus, the terminal voltage of the battery is 14 V, while its Emf is 16 V.
The loss of the voltage (= 2 V) is accounted for by the internal resistance1 Ω of the battery [2 A × 1 Ω = 2 V].