Consider the negative terminal of the 5 V battery is at zero volt and then we can write the potential of other nodes
Apply Kirchhoff rule of junction at the node having potential (V_{1} +12) Volt
$\frac {V_1 +12 -0}{4} + \frac {V_1 -0}{3} + \frac {V_1 +12 -5}{2} =0$
Or V_{1}=-6V
So current in 4 Ω=(V_{1}+12)/4=1.5 A
Current in 3 Ω=6/3=2A
Current in 2 Ω=(6-5)/2=.5 A
Potential difference across 4 Ω=6V
Three resistors of 1 Ω,2 Ω,3 Ω are connected to form a triangle DEF as shown in below figure.
Since the resistance along DEF are in series
So R=1+2=3 Ω
Now DEF is parallel to DF
So Net resistance
$R= \frac {3X3}{3+3} = 1.5 \Omega $
Let I be the current in the battery then
We know that
$I = \frac {E} {r+R}$
So
$I=\frac {3} {1+1.5} =1.2 A
Let I_{1} current flows along the path DF so I-I_{1} current will flow in the path DEF
PD across DF=I_{1}*3
PD across DEF=(I-I_{1})*3
Since PD will be same across the both
I_{1}=I/2=.6 A
So the current along DF=.6 A
And current across DEF=.6 A
Potential difference across the cell
= I_{1}*3
=.6*3=1.8 V
Potential difference along the DE
= I_{1}*1=.6 V
Let r be the internal resistance of the battery. Let G be the resistance of the galvanometer. Let E be the emf of each battery
When the batteries are connected in series
Total EMF=2E and Total internal resistance=2r
So the current in the circuit will be
$I =\frac {3} {G+2+2r}$
Given I=1A Therefore
$1=\frac {3} {G+2+2r}$
G+2=(3-2r) ---(1)
When the batteries are connected in parallel, the total emf E=1.5V and the total internal resistance =r/2.Hence the current in the circuit will be
$I^' =\frac {1.5}{G+2+ \frac {r}{2} } $
Given I^{’}=.6 A
Therefore
$.6= \frac {1.5}{G+2+ \frac {r}{2} } $
Or G+2=(2.5-r/2) --(2)
From equation 1 and 2
3-2r=2.5-r/2
Or r=1/3 Ω
Now from equation 1
G+2=3-(2/3)
G=1-(2/3)
=1/3 Ω
The Total resistance of the circuit before connecting the voltmeter
=1+2+3=6 ohm
The current in the circuit would be
I=10/6
The voltage across B resistors will be
=iR= (10/6)*2=10/3
This is the voltage to be measured
Now when the voltmeter is connected across the resistor B,The voltage will be parallel to the resistor .Since the voltmeter has internal résistance, The effective resistance across the resistors B = 2*10/(2+10)=5/3
Total resistance to the battery=1+5/3+3=17/3
Current =3E/17
Now voltage across the resistor B
=iR=(3E/17)(5/3)=5E/17=50/17
This will be the reading in the voltmeter
The error in reading =(10/3) –(50/17) =20/51
% error in reading= $\frac {20/51}{10/3} X 100 $
=11.76%
From parallel group of cell we know that
$E=\frac {E_1 r_1 + E_2 r_2}{r_1 + r+2} =E $
$r= \frac {r_1 r_2}{r_1 + r_2} = r/2 $
Now current
$I= \frac {E}{ \frac {r}{2} + R} = \frac {2E}{ r+2R}$
Power dissipated in the Resistance R=i^{2}R
$=\frac {(2E)^2 R}{ (r+2R)^2} = \frac {4E^2 R}{(r-2R)^2 + 8rR}$
For Power to be maximum the denominator should be minimum i.e (r-2R)^{2}=0
R=r/2
Therefore,
$P_max = \frac {4E^2 (r/2)}{(r+r)^2}=\frac {E^2}{2r}$
Let i_{1} and i_{2} be the current flowing from the cells of 1 and 2 volt respectively.
Applying the K rchhoff‘s first law at junction C ,the current in the 2 Ω will be i_{1}+i_{2}
Applying the Kirchhoff second law of closed mesh ADCA ,we have
i_{1}*1+(i_{1}+i_{2})*2=1
3i_{1}+2i_{2}=1 -(1)
Similarly for the closed mesh ABCA,we have
i_{2}*1+(i_{1}+i_{2})*2=2
2i_{1}+3i_{2}=2 -(2)
Solving equation (1) and (2),we have
I_{1}=-.2A
I_{2}=.8 A
The current in the 2 Ω resistors = i_{1}+i_{2}=.6 A
Suppose the EMF of the cell is E and internal resistance is r
Now we know
V=E-iR
When i=.5 A V=1.8
1.8=E-.5r -(1)
When i=1, V=1.6
1.6=E-r --- (2)
Solving equation (1) and (2)
E=2 volt and r=.4 ohm
Power is given by
P=VI
Given P=1000 and V=120
So 1000=120I
Or I=8.333 A
Now from Ohm’s law
V=IR
Or R=V/I
=14.4 ohm
The electric energy generated is given by
=Pt
Given P=600 W
So EE=600t=.600t KJ
The heat required to warm the water
=mc =.26*4.184*80=87.02 kJ
Now
Heat required to warm up the water=Electric energy generated by the heater
.600t=87.02
Or t=145 sec