It is an arrangement of four resistances used to measure one of them, in terms of the other three as shown in Fig. below.
For simplicity, we assume that the cell has no internal resistance. In general, there will be currents flowing across all the resistors as well as a current \(I_{ g }\) through G. Of special interest, is the case of a balanced bridge where the resistors are such that \(I_{ g }=0\). We can easily get the balance condition, such that there is no current through G. In this case, the Kirchhoff’s junction rule applied to junctions D and B (see the figure)
immediately gives us the relations \(I_1=I_3\) and \(I_2=I_4\). Next, we apply Kirchhoff’s loop rule to closed loops ADBA and CBDC. The first loop gives
\(
-I_1 R_1+0+I_2 R_2=0 \quad\left(I_{ g }=0\right) \dots(i)
\)
and the second loop gives, upon using \(I_3=I_1, I_4=I_2\)
\(
I_2 R_4+0-I_1 R_3=0 \dots(ii)
\)
From Eq. (i), we obtain,
\(
\frac{I_1}{I_2}=\frac{R_2}{R_1}
\)
whereas from Eq. (ii), we obtain,
\(
\frac{I_1}{I_2}=\frac{R_4}{R_3}
\)
Hence, we obtain the condition
\(
\frac{ R _2}{ R _1}=\frac{ R _4}{ R _3}
\)
This last equation relating the four resistors is called the balance condition for the galvanometer to give zero or null deflection. On mutually changing the position of cell and galvanometer, this condition will not change.
Different forms of Wheatstone bridge
Example 76: The four arms of a Wheatstone bridge (Fig. below) have the following resistances:
\(
AB =100 \Omega, BC =10 \Omega, CD =5 \Omega \text {, and } DA =60 \Omega \text {. }
\)
A galvanometer of \(15 \Omega\) resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.
Solution: Considering the mesh BADB, we have
\(
\begin{array}{ll}
& 100 I_1+15 I_g-60 I_2=0 \\
\text { or } \quad & 20 I_1+3 I_g-12 I_2=0 \dots(i)
\end{array}
\)
Considering the mesh BCDB, we have
\(
\begin{aligned}
& 10\left(I_1-I_g\right)-15 I_g-5\left(I_2+I_g\right)=0 \\
& 10 I_1-30 I_g-5 I_2=0 \\
& 2 I_1-6 I_g-I_2=0 \dots(ii)
\end{aligned}
\)
Considering the mesh ADCEA,
\(
\begin{aligned}
& 60 I_2+5\left(I_2+I_g\right)=10 \\
& 65 I_2+5 I_g=10 \\
& 13 I_2+I_g=2 \dots(iii)
\end{aligned}
\)
Multiplying Eq. (ii) by 10
\(
20 I_1-60 I_g-10 I_2=0 \dots(iv)
\)
From Eqs. (iv) and (i) we have
\(
\begin{aligned}
& 63 I_g-2 I_2=0 \\
& I_2=31.5 I_g \dots(v)
\end{aligned}
\)
Substituting the value of \(I_2\) into Eq. [iii)], we get
\(
\begin{aligned}
& 13\left(31.5 I_g\right)+I_g=2 \\
& 410.5 I_g=2 \\
& I_g=4.87 mA.
\end{aligned}
\)
Example 77: Find (a) Equivalent resistance (b) and current in each resistance
Solution: (a) \(R _{ eq }=\left(\frac{1}{16}+\frac{1}{8}+\frac{1}{16}\right)^{-1}+1=5 \Omega\)
(b) \(i=\frac{60}{4+1}=12 A\)
Hence 12 A will flow through the cell.
By using current distribution law.
Current in resistance \(10 \Omega\) and \(6 \Omega=3 A\)
Current in resistance \(5 \Omega\) and \(3 \Omega=6 A\)
Current in resistance \(20 \Omega=0\)
Current in resistance \(16 \Omega=3 A\)
Example 78: Find the equivalent resistance between A and B
Solution: This arrangement can be modified as shown in the figure since it is a balanced Wheatstone bridge
As it is a balanced bridge, the current through the arm CD is 0. So, \(R _{ eq }=\frac{2 R \times 2 R}{2 R+2 R}= R\)
Example 79: Determine the value of \(R\) in the circuit shown in the figure, when the current is zero in the branch CD.
Solution: As the bridge in balance condition \(5R=200\), \(R=40 \Omega\)
Example 80: Calculate the current drawn from the battery by the network of resistors shown in the figure.
Solution: The given network is equivalent to the circuit shown in the figure below.
Equivalent resistance, \(R_{ eq }=\frac{6 \times 3}{6+3}=2 \Omega\)
\(\therefore\) Current drawn from the battery,
\(
I=\frac{V}{R_{\text {eq }}}=\frac{4}{2}=2 A
\)
Measuring Instruments for Current and Voltage
There are various instruments like galvanometer, ammeter, and voltmeter which can be used to detect current and voltage, depending on the range.
Galvanometer
It is an instrument used to detect small current passing through it by showing deflection. It can be converted into voltmeter (for measuring voltage) and ammeter (for measuring current).
Ammeter
It is an instrument used to measure current and is always connected in series with the circuit element through which the current is to be measured. The smaller the resistance of an ammeter, more accurate will be its reading, as it will not change the circuit current. An ammeter is said to be ideal if its resistance \(r\) is zero.
Conversion of galvanometer into ammeter
A galvanometer can be converted into an ammeter by connecting a low resistance (called shunt \(S\) ) in parallel to the galvanometer of resistance \(G\).
Hence, only a small amount of current pass through the galvanometer and the remaining will pass through the shunt.
\(G\) and \(S\) are parallel and hence have equal potential difference,
i.e. \(i_g G=\left(i-i_g\right) S\).
Where \(\quad i=\) Range of ammeter
\(i_{ g }=\) Current required for full scale deflection of galvanometer.
\(G=\) Resistance of galvanometer coil.
\(\therefore\) Required shunt resistance, \(S=\frac{i_g}{\left(i-i_g\right)} G\)
Example 81: What shunt resistance is required to make the \(1 mA, 20 \Omega\) galvanometer into an ammeter with a range of 0 to 50 mA?
Solution: Given, \(i_g=1 mA=10^{-3} A, \quad G=20 \Omega\)
\(
i=50 \times 10^{-3} A
\)
Substituting in \(S=\left(\frac{i_g}{i-i_g}\right) G\)
\(
\begin{aligned}
& =\frac{\left(10^{-3}\right)(20)}{\left(50 \times 10^{-3}\right)-\left(10^{-3}\right)} \\
& =0.408 \Omega
\end{aligned}
\)
Example 82: What is the value of the shunt which passes \(10 \%\) of the main current through a galvanometer of 99 ohm?
Solution:
\(\begin{aligned} \text { As in figure } & R _{ g } I _{ g }=\left( I – I _{ g }\right) S \\ \Rightarrow & 99 \times \frac{I}{10}=\left(I-\frac{I}{10}\right) \times S \\ \Rightarrow & S =11 \Omega .\end{aligned}\)
Example 83: Find the current in the circuit (a) & (b) and also determine the percentage error in measuring the current through an ammeter.
Solution:
\(
\begin{array}{ll}
\ln A & i =\frac{10}{2}=5 A \\
\ln B & i^{\prime} =\frac{10}{2.5}=4 A
\end{array}
\)
Percentage error is \(=\frac{i-i^{\prime}}{i} \times 100=20 \%\)
Here we see that due to ammeter the current has reduced. A good ammeter has very low resistance as compared with other resistors so that due to its presence in the circuit the current is not affected.
Example 84: Find the reading of the ammeter. Is this the current through \(6 \Omega\)?
Solution:
\(
R _{ eq }=\frac{3 \times 6}{3+6}+1=3 \Omega
\)
Current through battery
\(
I =\frac{18}{3}=6 A
\)
So, current through ammeter
\(
=6 \times \frac{6}{9}=4 A
\)
No, it is not the current through the \(6 \Omega\) resistor.
An ideal ammeter is equivalent to zero resistance wire for calculation of potential difference across it is zero.
Voltmeter
It is an instrument used to measure potential difference and is always connected in parallel with the circuit element across which the potential difference is to be measured. The greater the resistance of the voltmeter, more accurate will be its reading, as only a small amount of current pass through it, by not changing the circuit current. A voltmeter is said to be ideal if its resistance is infinite.
Conversion of galvanometer into voltmeter
A galvanometer can be converted into a voltmeter by connecting a large resistance \(R\) in series with the galvanometer.
According to Ohm’s law, \(V=i_g(G+R)\) or required resistance, \(\quad R=\frac{V}{i_g}-G\)
Example 85: How can we make a galvanometer with \(G=20 \Omega\) and \(i_g=1 mA\) into a voltmeter with a maximum range of 10 V?
Solution: Using, \(R=\frac{V}{i_g}-G\)
We have, \(R=\frac{10}{10^{-3}}-20=9980 \Omega\)
Thus, a resistance of \(9980 \Omega\) is to be connected in series with the galvanometer to convert it into the voltmeter of the desired range.
Example 86: A galvanometer has a resistance of \(G\) ohm and range of \(V\) volt. Calculate the resistance to be used in series with it to extend its range to \(nV\) volt.
Solution: Full scale current \(i _{ g }=\frac{ V }{ G }\) to change its range
\(
V =\left( G + R _{ s }\right) i _{ g } \Rightarrow nV =\left( G + R _{ s }\right) \frac{ V }{ G } \Rightarrow R _{ s }= G ( n -1)
\)
Current sensitivity
The ratio of deflection to the current i.e. deflection per unit current is called current sensitivity (C.S.) of the galvanometer \(\operatorname{CS}=\frac{\theta}{ I }\)
Example 87: A galvanometer with a scale divided into 100 equal divisions, has a current sensitivity of 10 division per mA and voltage sensitivity of 2 division per mV. What adoptions are required to use it (a) to read 5 A full scale and (b) 1 division per volt?
Solution: Full scale deflection current \(i_g=\frac{\theta}{ cs }=\frac{100}{10} mA=10 ~mA\)
Full scale deflection voltage \(V _{ g }=\frac{\theta}{ VS }=\frac{100}{2} mv =50 ~mv\)
So galvanometer resistance \(G =\frac{ V _{ g }}{ i _{ g }}=\frac{50 mV }{10 mA}=5 \Omega\)
(a) to convert the galvanometer into an ammeter of range 5 A , a resistance of value \(S \Omega\) is connected in parallel with it such that
\(
\begin{aligned}
& \left( I – i _{ g }\right) S = i _{ g } G \\
& (5-0.01) S =0.01 \times 5 \\
& S=\frac{5}{499} \cong 0.01 \Omega
\end{aligned}
\)
(b) To convert the galvanometer into a voltmeter which reads 1 division per volt, i.e. of range 100 V , \(V = i _{ g }( R + G )\)
\(
\begin{aligned}
100 & =10 \times 10^{-3}( R +5) \\
R & =10000-5 \\
R & =9995 \Omega \cong 9.995 k \Omega
\end{aligned}
\)
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