# NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 3 (Part IV)

NCERT Exemplar Solutions for CBSE Class 12 Physics – Chapter 3 (Current Electricity) are available here. In this article, you will find solutions for short answer type questions (i.e. question number 3.22 to question number 3.27). Solutions of question number 3.1 to 3.21 are already available in **Part I**, **Part II **and **Part III**. Solutions of the remaining questions will be available in next part. These questions and solutions are important for CBSE Class 12th Physics board exam 2018 and other engineering & medical entrance exams like NEET, WBJEE, UPSEE, JEE Main etc.

*NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 3 (from question number 3.22 to 3.27) are given below:*

**Question 3.22:** First a set of *n *equal resistors of *R *each are connected in series to a battery of emf *E* and internal resistance* R*. A current *I* is observed to flow. Then the *n* resistors are connected in parallel to the same battery. It is observed that the current is increased 10 times. What is ‘*n*’?

**S****olution 3.22:**

*I*** =** *E */ (*R* + *nR*) [In case of series combination of resistor]

*I’* = *E* / (*R* + *R*/*n*) [In case of parallel combination of resistor]

According to the question,

*I*’ = 10 *I*

⇒ *E* / (*R* + *R*/*n*) = 10*E */ (*R* + *nR*)

⇒ *n* = 10.

**CBSE Class 12 Physics Syllabus 2017 – 2018**

**Question 3.23: **Let there be *n *resistors *R*_{1}...........*R*_{n} with *R*_{max} = max (*R*_{1}......... *R*_{n}) and R_{min} = min {*R*_{1}..... *R*_{n}. Show that when they are connected in parallel, the resultant resistance *R*_{P }< *R*_{min} and when they are connected in series, the resultant resistance *R*_{S} > *R*_{max}. Interpret the result physically.

**Solution 3.23:**

(1/*R*_{P}) = (1/*R*_{1}) +…+ (1/*R _{n}*),

Now, *R*_{min}/*R*_{P} = (*R*_{min}/*R*_{1}) + (*R*_{min}/*R*_{2}) + …+ (*R*_{min}/*R*_{n}) > 1 and R_{S} = R_{1} + ...... + R_{n} ≥ R_{max}.

In Figure (b),

*R*_{min} provides an equivalent route as in Figure (a) for current.

But, in addition there are (*n *– 1) routes by the remaining (*n *– 1) resistors.

Current in Figure (b) > current in Figure (a).

Effective Resistance in Fig. (b) < *R*_{min}. Second circuit evidently affords a greater resistance. You can use Fig. (c) and (d) and prove *R*_{s} > *R*_{max}.

**Question 3.24:** The circuit in Fig 3.6 shows two cells connected in opposition to each other. Cell E_{1} is of emf 6V and internal resistance 2Ω; the cell E_{2} is of emf 4V and internal resistance 8Ω. Find the potential difference between the points A and B.

**Solution 3.24:**

In the above arrangement, effective resistance = 2 Ω + 8 Ω = 10 Ω and effective emf of two cells = 6 - 4 = 2 V, so the electric current is, I = (6 ‒ 4)/(2 + 8) = 0.2 Ampere.

Here, E_{1} > E_{2}, therefore current will be along anti-clockwise direction.

As, current flows from higher potential to lower potential, so, *V _{B}* >

*V*

_{A}Now applying Kirchoff’s law, we have, *V _{B }*

*-*

*4*

*V*

*-*

*(0.2)*

*´ 8*

*= V*

_{A}⇒* V _{B} *

*-*

*V*= 3.6V.

_{A}**Question 3.25:** Two cells of same emf E but internal resistance *r*_{1} and *r*_{2} are connected in series to an external resistor *R* (Fig 3.7). What should be the value of R so that the potential difference across the terminals of the first cell becomes zero.

**Solution 3.25:**

In the given arrangement,

Effective resistance = *R* + *r*_{1} + *r*_{2} and effective emf = *E* + *E* = 2 *E*.

Therefore, electric current in the circuit, *I* = (*E* + *E*)/(*R* + *r*_{1} + *r*_{2}) = (2*E*)/(*R* + *r*_{1} + *r*_{2})

The potential difference across the terminals of the first cell and putting it equal to zero.

⇒ *V*_{1} = *E* ‒ *Ir*_{1} = *E* ‒ (2*Er _{1}*)/(

*R*+

*r*

_{1}+

*r*

_{2}) = 0

⇒ *E* = (2*Er _{1}*)/(

*R*+

*r*

_{1}+

*r*

_{2})

⇒ 1 = (2*r _{1}*)/(

*R*+

*r*

_{1}+

*r*

_{2})

⇒ *R* = *r*_{1} ‒ *r*_{2}.

**Question 3.26:** Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1mm. Conductor B is a hollow tube of outer diameter 2mm and inner diameter 1mm. Find the ratio of resistance *R*_{A} to *R*_{B}.

**Solution 3.26:**

Resistance of a wire is given by, *R* = ρ (*l*/*A*), where *l* is length of the conductor and *A* is the area of the cross-section & ρ is a constant.

Resistance of first conductor, *R*_{A} = ρ [*l*/π (10^{‒3} × 0.5)^{2}]

Resistance of second conductor, *R*_{B} = ρ [*l*/π {(10^{‒3})^{2} ‒ (0.5×10^{‒3})^{2}]

Ratio, *R*_{A}/*R*_{B} = [π {(10^{‒3})^{2} ‒ (0.5×10^{‒3})^{2}]/ (10^{‒3} × 0.5)^{2}

⇒ *R*_{A}/*R*_{B} = 3:1.

**Question 3.27:** Suppose there is a circuit consisting of only resistances and batteries. Suppose one is to double (or increase it to n-times) all voltages and all resistances. Show that currents are unaltered. Do this for circuit of Example 3.7 in the NCERT Text Book for Class XII.

**Solution 3.27:**

After reducing the circuit, let *R*_{eff} be the internal resistance of the battery, the effective external resistance is *R* and the effective voltage of the battery is *V*_{eff}. The situation is shown in the figure given below:

The current thorough R is given by,

*I* = *V*_{eff}/(*R*_{eff} + *R*)

If all the resistances and the effective voltage are increased n-times, then we have

*V*_{eff} * ^{new}* =

*n V*

_{eff}and

*R*

_{eff}

^{new }=

*n R*

_{eff}also

*R*

^{new}=

*nR*

Then, the new current is given by

*I*’ = *nV*_{eff}/(*nR*_{eff} + *nR*) = (*V*_{eff})/(*R*_{efff} + *R*) = *I.*

Hence, the current remains same.

**NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 3: Current Electricity**