Multiple choice questions on EMF and Electric measurements(Electric Circuits)

Voltmeter when connected across any circuit element should not draw any current from it. If the voltmeter resistance is not high as compared to the resistance of the circuit element, across which it is connected, the measured voltage becomes less. Thus resistance of the voltmeter should be as large as possible.
Hence (c) is the correct answer

(i) Consider the circuit given in the question

Rate at which energy is dissipated in 20 Ω is
P=I²R
So we need to calculate the current through 20 Ω resistor. Applying Kirchoff’s loop rule in ABCDA we get equation
E₁ -6I₁ +10 I₂ =0
or
20 -6I₁ +10 I₂ =0  ----(1)
Applying Kirchoff’s loop rule in CDEFC we get equation
E₂ -10I₂ -20I₃  =0
24- 10I₂ -20I₃  =0        ---(2)
From Kirchoff’s junction rule at point C
I₁ + I₂ = I₃
putting this in equation 2 we find
24-30I₂ -20I₁ =0       -----(3)
Solving equation 1 and 3 for the value of I₁ we get
I₁=2.21A
Again from equation 1 we find
I₂=-0.67A here consider reversing the direction of current from the assumed one.
And finally
I₃=1.54A
Rate at which energy is dissipated in 20 Ω is
P=I²R=(1.54)²x20=47.2 Watt
Hence (d) is the correct answer
(ii) I₂ is the current through 10Ω resistor
Hence (b) is the correct answer
(iii) Power of battery is
P_emf=iE
Power supplied by battery E₁ = I₁E₁=44.2 Watt
Hence (a)  the correct answer

Applying Ohm’s law in section PQ
V=IR
The resistances, R and internal resistance of the voltmeter are connected in parallel
$\frac {I}{V} =\frac {1}{R}  + \frac {1}{R_v}$
or
$\frac {1}{R} = \frac {I}{V} - \frac {1}{R_v} = \frac {R_v I -V}{VR_v}$
Thus
$R=\frac {V}{I - \frac {V}{R_v}}$
putting the respective values and calculating we find R=20.13Ω
Hence (c) the correct answer

$H=\frac {V^2 t}{R}$
but
$R=\frac {\rho L}{A} =\frac {\rho L}{\pi r^2}$
Hence
$H=\frac {V^2 t r^2 \pi}{\rho L}
Hence (b) the correct answer

Rate at which heat is produced in the wire is
$\frac {heat}{time} =\frac {V^2}{R} = 2205 Js^-1$
Latent heat of ice=80c/gm=80x4.2J/gm
Hence rate of melting ice would be =$\frac {2205}{80X4.2} =6.56g/s$
Hence (c) the correct answer

(d) is the correct answer

When radius of the wire is changed , it will not affect the ratio of resistance of two arms AC and BC of the slide wire bridge.
Hence (d) is the correct answer

10 (b)
11.(d)
12 (a)
13 (a)
 

Given that current I enter the cube at point A and leaves at point G as shown below in the figure

Here we have
I₁=2I₂
And also from point rule we have I=3I₁
If E is the emf connected across the system then
E=IR_eq
Applying Kirchhoff’s law to any path between A and G , say ABFG , we get
I₁R+I₂R+I₁R=V=IR_eq
Putting I₁=2I₂ in above equation we get
5I₂R=IR_eq
Since I=3I₁ so we have I=6I₂
From this we have
5I₂R=6I₂R_eq
Or, R_eq=5R/6

Let I be the amount of current through the ammeter, I1 be the current due to left hand EMF and $I_2$ be the current due to right hand emf. Applying Kirchhoff's law we find
$E-Ir-I_1pR_0=0$
$E-Ir-I_2(1-p)R_0=0$
and
$I=I_1+I_2$
Solving for I we get
$I=\frac{E}{\left(r+R_0p-R_0p^2\right)} $

For maximum I denominator should be smallest which is for two values of p i.e., for p=0,1.

From the figure in the question we see that all the resistors are connected in the series combination. So equivalent resistance of the circuit is
$R=R_1+R_2+R_3 = 6$ ω

Total current in the circuit is I=V/R = 12/6 =2A

For finding potential difference between points A and E we first need to find the effective resistance between these two points.
$R_{AE} = 2+3=5$ ω
$V_{AE} = IR_{AE} = 10$ V

$V_E$ is the voltage across the resistor $R_3$ . So,
$V_E=R_3I = +2$ V above the ground

Power supplied is P=VI = 24W

First consider circuit A for finding value of lowest potential. Point of lower potential in circuit A is the point of maximum negative potential and its value is
V=IR
Where
I= 24/160 = .15 A is the current in the circuit.
So,
Lowest potential is
V=-0.15 × 100 = -15 V
That is 15 V below the ground.
Now potential at point B in circuit A = 0.15 × 60 =+9V

Now we find circuit current in circuit B
Net voltage= 12-6 = 6V
Total resistance in the circuit = 4+8 = 12 ω
And circuit current is, I =6/12 = .5A
Now potential at point B in circuit B = $V_A-(4 \times 0.5)$
Again potential at point A in circuit B =12-6=+6V which is positive w.r.t. ground
$V_B = 6-2= +4V$
P -> W ; Q -> Y ; R - >U ; S ->V ; T- >X

When both the switches are open resistances $R_1$, $R_2$ and $R_3$ are in series combination,So current through ammeter would be
$I_A= \frac {E}{R_1 + R_2 + R_3}$

Again with both Switches closed

Voltage at points a and b would be same and no current passes through resistance $R_3$. Now current through E and $R_2$
$I'=\frac {E}{R_2 + \frac {RR_1}{R_1 + R}}$
Current through Ammeter would be the current through resistor $R_1$. If i is current through resistor R then I' -i would be the current through $R_1$ and ammeter. So from Kirchoff's law
$(I' -i)R_1 = iR$
or
$i= \frac {I' R_1}{R+ R_1}$
So current through ammeter
$=I' -i = I' - \frac {I' R_1}{R+ R_1}= \frac {I'R}{R+R_1}$
or
$=\frac {RE}{(R+R_1)R_2 + RR_1}$
Now since Ammeter reports same reading before and after
$\frac {RE}{(R+R_1)R_2 + RR_1}=\frac {E}{R_1 + R_2 + R_3}$
$R(R_1 + R_2 + R_3)=(R+R_1)R_2 + RR_1$
$RR_1 + RR_2 + RR_3 = RR_2 + R_1 R_2 + RR_1$
or
$R= \frac {R_1}{R_2}{R_3} = \frac {200 \times 600}{100} = 1200$ ω

In charging of a capacitor we have
$q=q_0(1-e^{-\frac{t}{RC}})$ --- (1)
Where $q_0 = CV$
(a) $q_0=CV=\left(20\times{10}^{-6}F\right)\left(6V\right)=\ 120\mu C$
(b) Time constant is given by
$\tau=RC=\left(200\Omega\right)\left(20\times{10}^{-6}F\right)=4\times{10}^{-3}s$.
Now from equation 1 if charge reaches ${\frac{1}{2}q_0}$ in time t then
$ e^{-\frac{t}{RC}}=\frac{1}{2}$
Taking log on both the sides and solving for t we get
$t=t_{1/2}=2.8 \times 10^{-3} $ s
(c) $-\frac{t}{\tau}=ln 0.2$
or, $\ t=\tau\times1.6=4\times{10}^{-3}\times1.6=6.4\times{10}^{-3}\ $s

First when switch $S_2$ is open and $S_1$ is closed the capacitor $C_4$ is not included in the circuit and the effective circuit is shown below in the figure.


If battery provides a voltage of 90V then $V_1+V_2+V_3=V$ where $V_1$, $V_2$, and $V_3$ are voltages across the capacitors $C_1$, $C_2$, $C_3$ respectively.
Since all the capacitors are in series combination so charge on all the capacitor plates is same say, Q. Also given that
$C_1=C_2=C_3=C=10 \mu F$
$Q=C_1V_1= C_2V_2= C_3V_3$
$V_1=V_2=V_3=V'$
which gives ,
$3V'=V$ or $V'=90/3 = 30$V
Once the capacitors are charged this way, the potential difference across each capacitor will remain the same even after opening the switch $S_1$.

If we now open switch $S_1$ then battery is excluded from the circuit after charging of the capacitors $C_1$, $C_2$ and $C_3$. Now closing switch $S_2$ will bring $C_4$ in the circuit and effective circuit would be


Capacitors $C_1$ and $C_3$ will still carry the same charge as before and opening of switch $S_1$ leaves them isolated from any flow of charge. Charge Q originally on capacitor $C_2$ is distributed between capacitor $C_2$ and $C_4$. As $C_2=C_4$ so charge on each capacitor equals Q/2 and voltage across $C_2$ and $C_4$ would be equal to half the original voltage across C2. Thus
$V_1=V_3=30V$ , $V_2=V_4=15V$

Again with capacitors initially uncharged and switch $S_1$ open , closing $S_2$ does not provide any potential difference across any capacitor as battery is effectively disconnected.

Now we close $S_1$ with $S_2$ closed previously and circuit arrangement is as shown below in the figure.

Effective capacitance of parallel combination between capacitors $C_2$ and $C_4$ is
$C_{24} = C+C = 2C$
And the entire system has equivalent capacitance
$\frac{1}{C_{eq}}=\left(\frac{1}{C_1}+\frac{1}{C_{24}}+\frac{1}{C_3}\right)$
Or,
$C_{eq}=\frac{2}{5}C$
Thus, charge drawn from the battery onto each of $C_1$, $C_24$ and $C_3$ is,
$Q'=C_{eq}V={\frac{2}{5}CV}$
$V_1=V_3=\frac{Q'}{C}=\frac{2}{5}CV=36V$
$V_2=V_4=\frac{Q'}{2C}=\frac{1}{5}CV=18V$
Hence P ->Y ; Q -> Z ; R -> W ; S ->X

(b), (d)