Let E and r be the emf and internal resistance of each cell.
Case I When the cells are in series.
Total emf of cells = 14E
Total resistance of circuit = 82.6 + 14r
∴ Current in the circuit is given by
\(\frac{14 E}{82.6+14 r}=0.25 \mathrm{~A}\)
Case II When the cells are in parallel.
Total emf of cells = E
Total resistance of circuit = 0.053 \(+\frac{r}{14}\)
∴ Current in the circuit is given by
\(\frac{E}{0.053+\frac{r}{14}}=25 \mathrm{~A}\)
Dividing Eq. (i) by Eq. (ii), we get
\(14 \frac{\left(0.053+\frac{r}{14}\right)}{(82.6+14 r)}=10^{-2}\)
\(\Rightarrow \ 14 \times \frac{14 \times 0.053+r}{14} \times 10^{2}=82.6+14 r\)
⇒ 53 x 14 + 100r = 82.6 + 14r
Solving, we get
r = 0.097Ω ≈ 0.1 Ω
Substituting the value of r in Eq. (i), we get
E = 1.5V

(a) Presence of mind, Knowledge of subject, Concern for his grandmother, Empathy, Helping and caring, attitude
(b) Power = energy/time = \(\frac{200}{2\times10^{-3}}\)
= 100kW

Here,
v = 6 x 10¹⁵ s^-1;
e = 1.6 x 10^-19 C
Equivalent current I = ev
 = (1.6 x 10^-19) x 6 x 10¹⁵
 = 9.6 x 10^-4 A
 = 9.6 x 10^-4 x 10³ mA
= 0.96 mA

According to the question, the circuit can be drawn as shown in the figure
 
where ε = 2 V, and r = internal resistance of each cell
An equivalent circuit can be redraw as shown.
\( \frac{r}{4} =\left[\frac{E-V}{V}\right] \times R \)
\(=\left[\frac{2-1.6}{1.6}\right] \times 7.5 \)
\(r =7.5 \Omega \)

Applying Kirchhoffs second law in loop BADB,
2 - 2i₁ - i₁ - 1 - 2 (i₁ - i₂) = 0      ......... (i)
Similarly; applying Kirchhoffs second law in loop BDCB,
2 (i₁ - i₂) + 3 - 3i₂ - i₂ - 1 = 0        ........(ii)
Solving Eqs. (i) and (ii), we get
\(i_{1}=\frac{5}{13}, i_{2}=\frac{6}{13}\)
and  \(i_{1}-i_{2}=-\frac{1}{13}\)
(i) Potential difference between B and D,.
VB + 2(i₁ - i₂) = V_D
∴ \(V_{B}-V_{D}=-2\left(i_{1}-i_{2}\right)=\frac{2}{13} \mathrm{~V}\)
(ii) V_G = E_G - i₂r_G = 3 - \(\frac{6}{13} \times 3=\frac{21}{13} \mathrm{~V}\)
\(V_{H}=E_{H}+i_{2} r_{H}=1+\frac{6}{13} \times 1=\frac{19}{13} \mathrm{~V}\)

Working principle: When constant current flows through a wire of uniform cross section then potential difference across the wire is directly proportional to the length. V a l With key K₂ open, balance is obtained at length I₁ (AN₁). Then,

When key K₂ is closed, the cell sends a current (I) through the resistance box (RB). If V is the terminal potential difference of the cell and balance is obtained at length l₂ (AN₂).
\(V=\varphi l_2\)
\(\Rightarrow\ \varepsilon /V=l_2/l_2\)
But  \(\frac{\varepsilon}{V}\) =\(\frac{I(R+r)}{IR}=(1+\frac{r}{R})\)
\(\therefore\ \ (1+\frac{r}{R})=\frac{l_1}{l_2}\)
\(\Rightarrow \ r=\frac{(l_1-l_2)}{l_2}R\)

(a) Principle of potentiometer:
The potential drop across the length of a steady current carrying wire of uniform cross-section is proportional to the length of the wire.
(i) We use a long wire to have a lower value of potential gradient (i.e. a lower 'least count' or greater sensitivity of the potentiometer.
(ii) The area of cross-section has to be uniform to get a 'uniform wire' as per the principle of the potentiometer.
(iii) The emf of the driving cell has to be greater than the emf of the primary cells as otherwise no balance point would be obtained.
ஃ The required graph is as shown below

The circuit can be redrawn as

Let the current through the various arms of the network are as shown above. Applying Kirchhoff's second rule to close mesh ABCDA, we have
4 - 2I₁ - 6I₁ - 8(I₁ + I₂) = 0
16I₁ + 8I₂ = 4    ............(i)
and for closed mesh CDEFC, we have
-8(I₁ + I₂) - 4I₂ + 6 - 1I₂ = 0
8I₁ + 13I₂ = 6  ............(ii)             
Solving Eqs. (i) and (ii), we get
\({ I }_{ 1 }=\frac { 1 }{ 36 } A \ and \ I_{ 2 }=\frac { 4 }{ 9 } A\)  
Potential difference across 8\(\Omega \) resistance
= (I₁ + I₂)8 \(=\left( \frac { 1 }{ 36 } +\frac { 4 }{ 9 } \right) \times 8=\frac { 17 }{ 36 } \times 8=3.78V\)

Let R be external resistance in series with the cell of emf E and internal resistance \(\gamma \). The current in circuit is \(I=\frac { E }{ R+r } \)
Case I: I = 05A, R = 12Ω, then
\(0.5=\frac{E}{12+r}\)
⇒ E = 05 (12 + r)
⇒ E = 6.0 +  05r  ......(i)
Casell : I = 0.25A,R = 25Ω ,then
\(0.25=\frac{E}{25+r}\)
⇒ E = 0.25 (25 + r)
⇒ E = 6.25 + 0.25r  ......(ii)
From Eqs. (i) and (ii), we get
6.0 + 05 r = 6.25 + 0.25 r
⇒ r = 1Ω
From Eq. (i), we get
E = 6.0 + 05 x (1) = 65V
Hence, (i) internal resistance of the cell is 1 Ω.
(ii) emf of the cell is 6.5 V.

Let the bridge be balanced when additional resistance x is put in series with S.
Then, (S + X)  \(=\frac{Q}{P} R\)
or  \(x=\frac{Q}{P} R-S=\frac{8}{7} \times 12-7=6.72 \Omega\)

I = ev
= (1.6 x 10^-19) x (6.8 x 10¹⁵)
= 1.088 x 10^-3 A

(b) current density \(j\left( \frac { I }{ A } \right) \)is also directed along E and the relation is:
\( j=\sigma E\)

The plot of V versus I is a straight line for materials that obey Ohm's law. So, from the figure, material P obeys Ohm's law.

We know that, V = E - Ir
∴ Current in the circuit \(I=\frac{E-V}{r}\)
If the value of V is small, for high value of current I, then the internal resistance r should be small as \(I \propto \frac{1}{r} .\)

All the three resistances are in parallel.
Therefore, \(\frac{1}{R_{\mathrm{eq}}}=\frac{1}{2 R}+\frac{1}{2 R}+\frac{1}{R}=\frac{2}{R}\)
\(\therefore \ R_{\mathrm{eq}}=\frac{R}{2}\)

a) in parallel, power dissipation \(\alpha\) 1/R 
Therefore 3\(\Omega \) wire will dissipate more power
b) In series, power dissipation \(\alpha\) R
Therefore 9\(\Omega \) wire will dissipate more power

When K₁ is closed and K₂ is open, then only the cell connected in upper part branch will work. When K₂ is closed and K₁ is open, then only the cell connected in lower branch will work.
(i) K₁ ⟶  closed, K₂ ⟶ open

Suppose null point occurs at J.
Apply KVl in smaller loop
E - IR = 0     ......(i)
where, R = resistance
E = IR ⇒ I = E/R
As, X increases the value of resistance R. So, current in the circuit (wire) decreases. Hence, R will be increased. Then I will decrease. We can say, as X increases the value of R, null point decrease.
(ii) K₂ ⟶ open, K₁ ⟶ closed
Then the circuit will be same as shown earlier. We see that resistance S is not involved in the circuit because K₂ is open.

So, from Eq (i), we get
E = RI
⇒ I = E/R
Here, R does not depend on the value resistance S.
So, R null point is not affected by decreasing the value of resistance S.

In series combination, the same current flows through each bulb. As resistance
\(R=\frac{V^2}{P} \text { or } R \propto \frac{1}{P}.\) Therefore, the resistance of 40 W bulb is highest for the give three bulbs. Heat produced \(\mathrm{H}=\mathrm{I}^2 \mathrm{Rt} \text { or } \mathrm{H} \propto \mathrm{R}\) so maximum heat is produced in 40 W lamp will glow more brightly than the other two bulbs.

Slide wire bridge and post office box.

This impression is misleading. In fact, there is no special attractive force that keeps a person stuck with a high power line while touching that wire, whereas a current of few milliampere is enough to disorganise our nervous system. As a result of it, the affected person loses temporarily his ability to exercise his nervous control to get himself free from the high power line.