In an experiment with a potentiometer,
Question: In an experiment with a potentiometer, VB= 10V. R is adjusted to be 50 Ω. A student wanting to measure voltage E1of a battery finds no null point possible. He then diminishes R to 10 Ω and is able to locate the null point on the last segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case. Solution: Equivalent resistance of the potentiometer = 50 Ohm + R Equivalent voltage across the potentiometer...
Read More →A room has AC run for 5 hours a day at a voltage of 220V.
Question: A room has AC run for 5 hours a day at a voltage of 220V. The wiring of the room consists of Cu of 1 mm radius and a length of 10 m. Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions? Solution: Power consumption in a day = 10 units Power consumption per hour = 2 units Power consumption = 2 units = 2 kW = 2000 J/s Power consumption in resistors, P = VI Whic...
Read More →Two cells of voltage 10V and 2V and internal resistances
Question: Two cells of voltage 10V and 2V and internal resistances 10Ω and 5Ω respectively are connected in parallel with the positive end of the 10V battery connected to the negative pole of 2V battery. Find the effective voltage and effective resistance of the combination. Solution: Kirchhoffs law is applied at c, I1 = I + I2 Kirchhoffs law is applied at efbae, 10 IR 10I2 = 0 10 = IR + 10I1 Kirchhoffs law is applied at cbadc, -2-IR+5I2 = 0 2 = 5I2- RI I2 = I1 I Substituting the values we get E...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{x}{3 x^{4}-18 x^{2}+11} d x$ Solution: Let $I=\int \frac{x}{3 x^{4}-18 x^{2}+11} d x$ Let $\mathrm{x}^{2}=\mathrm{t} \ldots .$ (i) $\Rightarrow 2 x d x=d t$ $I=\frac{1}{6} \int \frac{1}{t^{2}-6 t+\frac{11}{3}} d t$ $=\frac{1}{6} \int \frac{1}{t^{2}-2 t(3)+(3)^{2}-(3)^{2}+11} d t$ $=\frac{1}{6} \int \frac{1}{(t-3)^{2}-\frac{16}{3}} d t$ Put $t-3=u$.........(ii) $\Rightarrow \mathrm{dt}=\mathrm{du}$ $I=\frac{1}{6} \int \frac{1}{(u)^{2}-\left(...
Read More →Suppose there is a circuit consisting of only resistances
Question: Suppose there is a circuit consisting of only resistances and batteries and we have to double all voltages and all resistances. Show that currents are unaltered. Solution: Reff is the equivalent internal resistance of the battery Veff is the equivalent voltage of the battery Using Ohms law, I = Veff/(Reff + R) When the resistance and effective voltage are increased by n-times, Veff = nVeff Reff = n Reff Rnew = nR Then current is given as I = bVeff/nReff + nR = Veff/Reff + R = I...
Read More →Two conductors are made of the same material
Question: Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1 mm. Conductor B is a hollow tube of outer diameter 2 mm and inner diameter 1 mm. Find the ratio of resistance RAto RB. Solution: Resistance of wire R = l/A Where A is the cross-sectional area of the conductor L is the length of the conductor is the specific resistance RA= l/(10-3 0.5)2 RB= l/ [10-3)2 (0.5 10-3)2] Therefore, the ratio is RA/RB= 3:1...
Read More →Two cells of same emf E but internal resistance r1 and r2
Question: Two cells of same emf E but internal resistance r1and r2are connected in series to an external resistor R. What should be the value of R so that the potential difference across the terminals of the first cell becomes zero. Solution: Effective emf of two cells = E + E = 2E Effective resistance = R + r1 + r2 Electric current is given as I = 2E/R+r1+r2 Potential difference is given as V1 E Ir1= 0 Which f=gives R = r1 r2...
Read More →Find the truth set in case of each of the following open sentences defined on N:
Question: Find the truth set in case of each of the following open sentences defined on N: (i) $x+210$ (ii) $x+54$ (iii) $x+32$ Solution: The open sentence x + 2 10 is defined on N; the set of natural numbers. $\mathrm{N}:\{1,2,3,4 \ldots\}$ $x=1 \rightarrow x+2=310$ $x=2 \rightarrow x+2=410$ $x=3 \rightarrow x+2=510$ $x=4 \rightarrow x+2=610$ $x=5 \rightarrow x+2=710$ $x=6 \rightarrow x+2=810$ $x=7 \rightarrow x+2=910$ $x=8 \rightarrow x+2=10$ So, $\exists x \in N$, such that $x+210$ $x=\{1,2,3...
Read More →The circuit in the figure shows two cells connected
Question: The circuit in the figure shows two cells connected in opposition to each other. Cell E1is of emf 6V and internal resistance 2Ω; the cell E2is of emf 4V and internal resistance 8 Ω. Find the potential difference between the points A and B. Solution: Applying Ohms law, equivalent emf of the two cells = 6 4 = 2V Equivalent resistance = 2 + 8 = 10 Ω Electric current, I = 6-4/2+8 = 0.2A When the loop is considered in the anti-clockwise direction, E1 E2which means VB VA Therefore, VB VA= 3....
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{x}{x^{4}-x^{2}+1} d x$ Solution: Let $I=\int \frac{x}{x^{4}-x^{2}+1} d x$ Let $\mathrm{x}^{2}=\mathrm{t} \ldots \ldots \cdots$ (i) $\Rightarrow 2 x d x=d t$ $I=\frac{1}{2} \int \frac{1}{t^{2}-t+1} d t$ $=\frac{1}{2} \int \frac{1}{t^{2}-2 t\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1} d t$ $=\frac{1}{2} \int \frac{1}{\left(t-\frac{1}{2}\right)^{2}+\frac{3}{4}} d t$ Put $\mathrm{t}-1 / 2=\mathrm{u}$ ......
Read More →Let there be n resistors R1……..Rn with Rmax = max(R1……Rn)
Question: Let there be n resistors R1..Rn with Rmax = max(R1Rn) and Rmin = min(R1.Rn). Show that when they are connected in parallel, the resultant resistance Rp Rmin and when they are connected in series, the resultant resistance Rs Rmax. Interpret the result physically. Solution: Following is the parallel grouping of the resistance: Rp = 1/R1 + 1/R2 + . + 1/Rn Rmin/Rp = Rmin/R1 + Rmin/R2 + ..+Rmin/Rn The RHS is equal to 1 Therefore, Rp Rmin Following is the series grouping of the resistance: R...
Read More →Split each of the following into simple sentences and determine whether
Question: Split each of the following into simple sentences and determine whether it is true or false. Also, determine whether an inclusive or or exclusive or is used. (i) The sum of 3 and 7 is 10 or $11 .$ (ii) $(1+i)$ is a real or a complex number. (iii) Every quadratic equation has one or two real roots. (iv) You are wet when it rains, or you are in a river. (v) 24 is a multiple of 5 or $8 .$ (vi) Every integer is rational or irrational. (vii) For getting a driving license, you should have a ...
Read More →First, a set of n equal resistors of R each are connected
Question: 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? Solution: When the resistors are in series combination, the current is given as I = E/R+nR When the resistors are in parallel combination, current is given as 10I = E/(R+R/n) Solving the two equations, we g...
Read More →A cell of emf E and internal resistance r is connected
Question: A cell of emf E and internal resistance r is connected across an external resistance R. Plot a graph showing the variation of PD across R versus R. Solution: We know that $I=\frac{E}{R+r}$ and V=IR. So $V=\frac{E R}{R+r} \ldots($ (I) $V=\frac{E}{1+\frac{\gamma}{R}} \ldots$ (II) Here $\mathrm{E}, \mathrm{r}$ are constants. So $V \propto \frac{1}{1+\frac{\gamma}{R}}($ from II $)$ and $I$ oc $R$ (from I) With increase in $\mathrm{R}$, P.D. across $\mathrm{R}$ is increased upto maximum val...
Read More →Split each of the following into simple sentences and determine
Question: Split each of the following into simple sentences and determine whether it is true or false. (i) A line is straight and extends indefinitely in both the directions. (ii) A point occupies a position, and its location can be determined. (iii) The sand heats up quickly in the sun and does not cool down fast at night. (iv) 32 is divisible by 8 and 12 . (v) $x=1$ and $x=2$ are the roots of the equation $x^{2}-x-2=0$. (vi) 3 is rational, and $\sqrt{3}$ is irrational. (vii) All integers are r...
Read More →While doing an experiment with potentiometer
Question: While doing an experiment with potentiometer it was found that the deflection is one-sided and i) the deflection decreased while moving from one end A of the wire to the end B; ii) the deflection increased, while the jockey was moved towards the end B. (i) Which terminal +ve or ve of the cell E, is connected at X in case i) and how is E1 related to E? (ii) Which terminal of the cell E1 is connected at X in case ii)? Solution: (i) E is connected to the positive terminal of the cell E1 a...
Read More →AB is a potentiometer wire.
Question: AB is a potentiometer wire. If the value of R is increased, in which direction will the balance point J shift? Solution: When the value of R is increased, the potential difference across AB decreased and hence the potential gradient across AB decreases. This is given as E = kl Jockey shifts towards B when the balance length l increases....
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{x\left(x^{6}+1\right)} d x$ Solution: let I $=\int \frac{1}{x\left(x^{6}+1\right)} d x$ $=\int \frac{x^{5}}{x^{6}\left(x^{6}+1\right)} d x$ Let $x^{6}=t \ldots$ (i) $\Rightarrow 6 \mathrm{x}^{5} \mathrm{dx}=\mathrm{dt}$ $I=\frac{1}{6} \int \frac{1}{t(t+1)} d t$ $I=\frac{1}{6} \int\left(\frac{1}{t}-\frac{1}{t+1}\right) d t$ $I=\frac{1}{6}\left(\int \frac{1}{t} d t-\int \frac{1}{(t+1)} d t\right)$ $I=\frac{1}{6}(\log t-\log (t+1))+c$ $I=\f...
Read More →Power P is to be delivered to a device via transmission
Question: Power P is to be delivered to a device via transmission cables having resistance Rc. If V is the voltage across R and I the current through it, find the power wasted and how can it be reduced. Solution: Power consumed by the transmission lines is given as P = i2Rc Where Rc is the resistance of connecting cables Power is given as P = VI The transmission of power takes places either during low voltage and high current or during high voltage and low current....
Read More →Why are alloys used for making
Question: Why are alloys used for making standard resistance coils? Solution: Alloys are used in the making of the standard resistance coils because they have less temperature coefficient of resistance and the temperature sensitivity is also less....
Read More →For wiring in the home,
Question: For wiring in the home, one uses Cu wires or Al wires. What considerations are involved in this? Solution: The main considerations in the selection of the wires is the conductivity of the metal, cost of metal, and their availability....
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{x^{2}}{x^{6}+a^{6}} d x$ Solution: let $I=\int \frac{x^{2}}{x^{6}+a^{6}} d x$ $=\int \frac{x^{2}}{\left(x^{3}\right)^{2}+\left(a^{3}\right)^{2}} d x$ Let $\mathrm{x}^{3}=\mathrm{t} \ldots \ldots$ (i) $\Rightarrow 3 x^{2} d x=d t$ $I=\frac{1}{3} \int \frac{1}{t^{2}+\left(a^{3}\right)^{2}} d t$ $I=\frac{1}{3 a^{3}} \tan ^{-1} \frac{t}{a^{3}}+c$ $\left[\right.$ since, $\left.\int \frac{1}{\mathrm{x}^{2}+(\mathrm{a})^{2}} \mathrm{dx}=\frac{1}{\...
Read More →What is the advantage of using
Question: What is the advantage of using thick metallic strips to join wires in a potentiometer? Solution: The advantage of using thick metallic strips is that the resistance of these strips are negligible....
Read More →What are the advantages of the null-point method
Question: What are the advantages of the null-point method in a Wheatstone bridge? What additional measurements would be required to calculate R unknown by any other? Solution: The advantage of a null-point in the Wheatstone bridge is that the resistance of the galvanometer is not affected by the balance point. The R unknown is calculated by using Kirchhoffs rule....
Read More →The relaxation time τ is nearly independent
Question: The relaxation time is nearly independent of applied E field whereas it changes significantly with temperature T. First fact is responsible for Ohms law whereas the second fact leads to a variation of with temperature. Elaborate why? Solution: Relaxation time is the time interval between two successive collisions of the electrons. It is defined as = mean free path/rms velocity of electrons Usually, the drift velocity of the electrons is small because there is a frequent collision betwe...
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