Physics > Current Electricity > 6.0 Heating effect of current
Current Electricity
1.0 Introduction
2.0 Conduction of current in a metal
3.0 Ohm's law
3.1 Temperature dependence of resistance
3.2 Resistivities of different materials
3.3 Limitations of ohm's law
4.0 Combination of Resistors
5.0 Electromotive force $\left( \xi \right)$
6.0 Heating effect of current
7.0 Wheatstone bridge
8.0 Metre Bridge Or Slide wire bridge
9.0 Potentiometer
9.1 Comparison of emfs of two primary cells.
9.2 Determination of Internal resistance of a cell using potentiometer
10.0 Electrical devices
6.2 Kirchhoff's law
3.2 Resistivities of different materials
3.3 Limitations of ohm's law
9.2 Determination of Internal resistance of a cell using potentiometer
1. Kirchhoff's first law or junction law. It states that algebraic sum of currents meeting at a junction is zero i.e $\Sigma I = 0$
According to this law , there is no accumulation of electric current at any junction. So, current entering the junction is equal to current leaving the junction.
This law follows law of conservation of charge.
By convention, current which flows towards a junction is taken as positive and current which flow away from junction is taken as negative.
2. Kirchhoff's second law or loop law. It states that, in a closed circuit, the algebraic sum of all potential difference is zero. $$\Sigma V = 0 \Rightarrow \Sigma \xi = \Sigma IR$$
This law follows law of conservation of energy.
By convention, potential difference across a resistor is taken to be negative when we move in direction of flow of current. The emf is taken positive when we move from negative to positive terminal through the cell and vice versa.
Let us understand Kirchhoff's law better with the help of an example.
| Case 1 | Case 2 | |
| Direction |  |  |
From Kirchhoff's loop law,
Case 1: $$ \Rightarrow - {I_1}{R_1} - {I_2}{R_2} + {\xi _2} - {I_3}{R_3} + {I_4}{R_4} - {\xi _1} = 0$$
Case 2: $$ \Rightarrow {I_1}{R_1} + {\xi _1} - {I_4}{R_4} + {I_3}{R_3} - {\xi _2} + {I_2}{R_2} = 0$$
Question 16. Using Kirchhoff's law in the electrical network shown in the figure, calculate ${I_1},{I_2}$ and ${I_3}$.
Solution:
Applying kirchhoff's junction law at point $B$, $${I_1} + {I_3} = {I_2}$$
Applying kirchhoff's second law to loop $ABEDA$ we get, $$ \Rightarrow 12 - 5{I_1} - 2{I_2} = 0$$
Applying kirchhoff's second law to loop $BCFEB$ we get, $$ \Rightarrow 3{I_3} - 6 + 2{I_2} = 0$$
On solving above three equations using Cramer's rule as, $$\begin{equation} \begin{aligned} {I_1} - {I_2} + {I_3} = 0 \\ 5{I_1} + 2{I_2} + 0 = 12 \\ 0 + 2{I_2} + 3{I_3} = 6 \\\end{aligned} \end{equation} $$
\[\Delta = \left| {\begin{array}{c} 1&{ - 1}&1 \\ 5&2&0 \\ 0&2&3 \end{array}} \right| = 31\]
\[{\Delta _1} = \left| {\begin{array}{c} 0&{ - 1}&1 \\ {12}&2&0 \\ 6&2&3 \end{array}} \right| = 48\]
\[{\Delta _2} = \left| {\begin{array}{c} 1&0&1 \\ 5&{12}&0 \\ 0&6&3 \end{array}} \right| = 66\] \[{\Delta _3} = \left| {\begin{array}{c} 1&{ - 1}&0 \\ 5&2&{12} \\ 0&2&6 \end{array}} \right| = 18\]
So, $$\begin{equation} \begin{aligned} {I_1} = \frac{\Delta_1 }{{{\Delta}}} = \frac{{48}}{{31}}A \\ {I_2} = \frac{\Delta_2 }{{{\Delta}}} = \frac{{66}}{{31}}A \\ {I_3} = \frac{\Delta_3 }{{{\Delta}}} = \frac{{18}}{{31}}A \\\end{aligned} \end{equation} $$
Question 17. Determine the current in each branch of the network shown in figure.
Solution:
Let I, ${I_1}$, ${I_2}$, ${I_3}$ be the current as shown in figure.
Applying kirchhoff's second law to,
Loop $ABDA$,$$10{I_1} + 5{I_3} - 5{I_2} = 0$$
Loop $BCDB$,$$5\left( {{I_1} - {I_3}} \right) - 10\left( {{I_2} + {I_3}} \right) - 5{I_3} = 0$$
Loop $ADCFGA$,$$5{I_2} + 10\left( {{I_2} + {I_3}} \right) + 10\left( {{I_1} + {I_2}} \right) = 10$$
Solving equations we get, $$\begin{equation} \begin{aligned} 10{I_1} - 5{I_2} + 5{I_3} = 0 \\ 5{I_1} - 10{I_2} - 20{I_3} = 0 \\ 10{I_1} + 25{I_2} + 10{I_3} = 10 \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} {I_1} = \frac{4}{{17}}A \\ {I_2} = \frac{6}{{17}}A \\ {I_3} = \frac{{ - 2}}{{17}}A \\\end{aligned} \end{equation} $$
Therefore,
Current in branch $AB$ = ${I_1} = \frac{{4}}{{17}}A$
Current in branch $BC$ = ${I_1} - {I_3} = \frac{6}{{17}}A$
Current in branch $DC$ = ${I_2} + {I_3} = \frac{4}{{17}}A$
Current in branch $AD$ = ${I_2} = \frac{{6}}{{17}}A$
Current in branch $BD$ = ${I_3} = \frac{{-2}}{{17}}A$
Total current, $$I = {I_1} + {I_2} = \frac{{10}}{{17}}A$$
Question 18. In Circuit shown in the figure, Find the potential difference ${\phi _A} - {\phi _B}$ between the plates $A$ and $B$ of the capacitor $C$. The internal resistances of sources are negligible.
Solution:
Current will not flow through the capacitor. So, we can short point $C$ & $F$.
Therefore, the current will flow only in the outer loop $ABDEA$.
Let the current flowing through the circuit be $I$.
At steady State (equilibrium state) no current flows through the capacitor,
Applying kirchhoff's law to loop $ABDEA$, $$-IR - V + 2V - 2IR = 0$$
$$ \Rightarrow 3IR = V \Rightarrow I = \frac{V}{{3R}}$$
Again, applying kirchhoff's second law to loop $CDEFC$, $${\phi _A} - {\phi _B} - IR - V + V$$
$$ \Rightarrow {\phi _A} - {\phi _B} = IR = \frac{V}{3}$$
Question 19. In the given network shown, find the values of ${I_1},{I_2}$ and ${I_3}$.
Applying kirchhoff's first law, $${I_1} + {I_2} = {I_3}$$
Applying kirchhoff's second law to loop $ADBA$, $$\begin{equation} \begin{aligned} \Rightarrow - {I_1} - 1 - 2{I_2} + 2 - 2{I_1} = 0 \\ \Rightarrow 3{I_1} + 2{I_2} = 1 \\\end{aligned} \end{equation} $$
Again, applying kirchhoff's second law to loop $DCBD$,$$\begin{equation} \begin{aligned} \Rightarrow 3 - 3{I_3} - {I_3} - 1 + 2{I_2} = 0 \\ \Rightarrow 4{I_3} - 2{I_2} = 2 \\\end{aligned} \end{equation} $$
On solving below equations we get, $$\begin{equation} \begin{aligned} {I_1} + {I_2} = {I_3} \\ 3{I_1} + 2{I_2} = 1 \\ 4{I_3} - 2{I_2} = 2 \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} {I_1} = \frac{5}{{13}}A \\ {I_2} = \frac{{ - 1}}{{13}}A \\ {I_3} = \frac{6}{{13}}A \\\end{aligned} \end{equation} $$
