Combination of Resistances Numerical Class-12 Nootan ISC Physics Ch-5 Electric Resistance and Ohms Law. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Combination of Resistances Numerical Class-12 Nootan ISC Physics Ch-5 Electric Resistance and Ohms Law
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-5 | Electric Resistance and Ohm’s Law |
| Topics | Numericals on Combination of Resistances |
| Academic Session | 2025-2026 |
Combination of Resistances
Numerical Class-12 Nootan ISC Physics Ch-5 Electric Resistance and Ohms Law
Que-21. How can three resistances of 2 Ω, 3 Ω and 6 Ω be connected to give an equivalent resistance of 4 Ω?
Ans-21 Three resistors of resistances 2Ω, 3Ω, and 6Ω respectively.
Let 𝑅1 = 2Ω
⇒ 𝑅2 = 3Ω
⇒ 𝑅2 = 6Ω
To obtain a total resistance of 4Ω from three resistors of given resistances, Firstly, connect
the two resistors of 3Ω and 6Ω in parallel to get a total resistance of 2Ω which is less than the lowest individual resistance.
When R2 and R3 are connected in parallel with R1 in series we get
⇒ 1/𝑅 = 1/𝑅2 + 1/𝑅3
⇒ 1/𝑅 = 1/3 + 1/6
⇒ 1/𝑅 = 1/2
⇒ 𝑅 = 2Ω
⇒ Resistance in series = 𝑅+𝑅1 = 2+2 = 4Ω
Hence, the total resistance of the circuit is 4Ω.
Que-22. Join three 3 Ω resistances to obtain an effective resistance of 2 Ω.
Ans-22 Two in series and third in parallel of them.
Que-23. Α 5 Ω resistor is connected in series with a parallel combination of n resistors each of 6 Ω. The equivalent resistance is 7 Ω. Find n.
Ans-23 If n resistors are connected in parallel each having an equivalent resistance of RΩ then total equivalent resistance R/n Ω.
R = 6Ω
R eqi = 6/n Ω
R net = 5+(6/n)
{5+(6/n)} = 7
5n+6 = 7n
7n-5n = 6
n = 3.
Que-24: Three resistors are joined to form a triangle ABC. The resistances of the sides AB, BC and CA are 40 Ω, 60 Ω and 100 Ω respectively. Find the equivalent resistance between points A and B.
Ans-24 To calculate resistance between A and B, we can treat resistors between BC and AC to be in series,
There equivalent hence is,
R′ = R BC+R AC
R′ = 60+100 = 160Ω
Now this equivalent resistor R′ is in parallel with RAB.
Equivalent of R′ and RAB can be calculated using formula,
R EQ = {R′ (RAB)}/{RAB+R′}
⟹ R EQ = {160×40}/(160+40) = 32Ω.
Que-25: Four resistors, each 10 Ω are joined in the form of a square. Find the equivalent resistance between two diagonally-opposite corners.
Ans-25 Resistance in the path ADC = 10Ω+10Ω = 20Ω
Resistance in the path ABC = 10Ω+10Ω = 20Ω
Since the resistances in the paths ADC and ABC are in parallel. Resistance between A and C, i.e.,
1/R = 1/20 + 1/20 = 2/20 = 1/10 or R = 10Ω.
Que-26: Calculate the equivalent resistance between points A and B in each of the shown networks of resistors.
Ans-26 (a) Connecting wire from A to B one neglected so all the thrice resistance came into parallel
Net resistance
1/R = 1/10 + 1/10 + 1/10
1/R = 3/10 = R = 10/3 Ω.
(b) Two parallel from 10Ω to give 5Ω
again net resistance
= 10 + {(10×15)/(10+15)}
= 10 + (150/25)
= 10 + 6 = 16Ω.
Que-27: Calculate the equivalent resistance between points A and B in each of the following networks of resistors.
Ans-27 (a) This arrangement is a balanced Wheat Stone Bridge, therefore 7Ω resistance will become in effective
therefore, net resistance
= {(10×15)/(10+15)}
= 150/25 = 6Ω.
(b) It is also an example of Wheat Stone Bridge
∴ net resistance
= (6×3)/(6+3) = 2Ω.
(c) These three resistance are in parallel as connecting wires can be neglected
∴ net resistance
R = 6/3 = 2Ω.
Que-28: A wire whose resistance 0.1 Ω/cm, is bent in the form of a square ABCD of side 10 cm. An exactly similar wire is connected across the points B and D. What will be the equivalent resistance between the other two corners A and C of the square? If a 2 V battery be connected between A and C, then what will be the current drawn from the battery? If an ammeter be connected across B and D, what will it read?
Ans-28 Resistance of each arm = 0.1 x 10 = 1Ω
again resistance of arm BD = 0.1 x 40 = 4Ω
as in between B and D same wire is connected but it is balanced wheat stone bridge then for resistance of BD is neglected.
and net resistance = (1+1)/2 = 1Ω
again current i = V/R
= 2/1 = 2A
and BD has no current.
— : End of Combination of Resistances Numerical Class-12 Nootan ISC Physics Ch-5 Electric Resistance and Ohms Law. :–
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