Wheatstones Bridge Meter Bridge And Potentiometer L-10 Wheatstone's Bridge, Meter Bridge and Potentiometer
→ \rightarrow → → \rightarrow → Wheatstone's Bridge, Meter Bridge and Potentiometer → \rightarrow → → \rightarrow →
Wheatstones Bridge Meter Bridge And Potentiometer L-10 Rules
→ \rightarrow → → \rightarrow → Rules → \rightarrow → → \rightarrow →
Wheatstones Bridge Meter Bridge And Potentiometer L-10 Example
A B C D E F A
2 I " + I " + I " + 2 I " − I − I = 0 2 I "+I "+I "+2 I "-I-I=0 2 I " + I " + I " + 2 I " − I − I = 0
I " = I 3 ∣ I "=\frac{I}{3} \mid I " = 3 I ∣
I i n = 2 I + 2 I " I _{ in }=2 I+2 I " I in = 2 I + 2 I "
= 2 I + 2 3 I = 8 I 3 =2 I+\frac{2}{3} I=\frac{8 I}{3} = 2 I + 3 2 I = 3 8 I
V = 8 I 3 R e q ≡ 2 I R V=\frac{8 I}{3} R_{e q} \equiv 2 I R V = 3 8 I R e q ≡ 2 I R
R e q = 3 4 R_{e q}=\frac{3}{4} R e q = 4 3
Wheatstone's Bridge, Meter Bridge and Potentiometer → \rightarrow → → \rightarrow → Example → \rightarrow → → \rightarrow →
Wheatstones Bridge Meter Bridge And Potentiometer L-10 Examples
Right hand loop
− ( I 2 − I 3 ) .2 + ( I 2 + I 3 − I 1 ) × 2 + 5 = 0 -(I_2 - I_3).2 + (I_2+I_3-I_1) \times 2 + 5=0 − ( I 2 − I 3 ) .2 + ( I 2 + I 3 − I 1 ) × 2 + 5 = 0
2 I 1 + 4 I 2 − 4 I 3 = 5 2I_1 + 4 I_2 - 4 I_3 = 5 2 I 1 + 4 I 2 − 4 I 3 = 5
Upper Left Loop
− I 2 ⋅ 4 − ( I 2 + I 3 ) ⋅ 2 − I 1 + 10 = 0 -I_2 \cdot 4-\left(I_ 2+I_ 3\right) \cdot 2-I_1 +10=0 − I 2 ⋅ 4 − ( I 2 + I 3 ) ⋅ 2 − I 1 + 10 = 0
I 1 + 6 I 2 + 2 I 3 = 10 I_1+6 I_2+2 I_3=10 I 1 + 6 I 2 + 2 I 3 = 10
− ( I 1 − I 2 ) 4 + ( I 2 + I 3 − I 1 ) × 2 − I 1 ⋅ 1 = 0 -(I_1 - I_2)4 + (I_2+I_3-I_1) \times 2-I_1 \cdot 1=0 − ( I 1 − I 2 ) 4 + ( I 2 + I 3 − I 1 ) × 2 − I 1 ⋅ 1 = 0
7 I 1 + 6 I 2 + 2 I 3 = 10 7I_1 + 6 I_2 + 2 I_3 = 10 7 I 1 + 6 I 2 + 2 I 3 = 10
I 1 = 5 2 A I_1=\frac{5}{2} \mathrm{A} I 1 = 2 5 A I 2 = 5 8 A I_2=\frac{5}{8} \mathrm{A} I 2 = 8 5 A I 3 = 15 8 A I_3=\frac{15}{8} \mathrm{A} I 3 = 8 15 A
Rules → \rightarrow → → \rightarrow → Examples → \rightarrow → → \rightarrow →
Wheatstones Bridge Meter Bridge And Potentiometer L-10 Example
LOOP BCDAB
− 2 I 1 R − ( I 1 − I 2 ) R + I 2 R = 0 -2 I_1 R-\left(I_1-I_2\right) R+I_2 R=0 − 2 I 1 R − ( I 1 − I 2 ) R + I 2 R = 0
I 2 = 3 2 I 1 I_2=\frac{3}{2} I_1 I 2 = 2 3 I 1
ADFEXYA
− I 2 R − 2 I 1 R = V -I_2 R-2 I_1 R =V − I 2 R − 2 I 1 R = V I 1 = 2 7 V R I_1=\frac{2}{7} \frac{V}{R} I 1 = 7 2 R V
I 1 + I 2 = 5 4 ⋅ V R = V R e q I_1+I_2=\frac{5}{4} \cdot \frac{V}{R}=\frac{V}{R {eq}} I 1 + I 2 = 4 5 ⋅ R V = R e q V
∴ R e q = 7 5 R \therefore R_{eq}=\frac{7}{5} R ∴ R e q = 5 7 R
Example → \rightarrow → → \rightarrow → Example → \rightarrow → → \rightarrow →
Wheatstones Bridge Meter Bridge And Potentiometer L-10 Example
EFOE:(LOOP)
− I 3 R − I 3 R + ( I 2 − I 3 ) R = 0 -I_3 R-I_3 R+\left(I_2-I_3\right) R =0 − I 3 R − I 3 R + ( I 2 − I 3 ) R = 0
⇒ I 3 = I 2 / 3 \Rightarrow I_3=I_2 / 3 ⇒ I 3 = I 2 /3
EOAE :
− I 2 R − I 3 R + I 1 R = 0 -I_2 R-I_3 R+I_1 R=0 − I 2 R − I 3 R + I 1 R = 0
I 2 = 3 4 I 1 I_2=\frac{3}{4} I_1 I 2 = 4 3 I 1
I 1 + I 2 = 7 4 I 1 , I_1+I_2=\frac{7}{4} I_1, I 1 + I 2 = 4 7 I 1 ,
2 I 1 R = V 2 I_1 R=V 2 I 1 R = V
I 1 = V 2 R I_1=\frac{V}{2 R} I 1 = 2 R V
= 7 8 V R = V R e q = \frac{7}{8}\frac{V}{R}=\frac{V}{R_{eq}} = 8 7 R V = R e q V
R e q = 8 7 ? R_{eq}=\frac{8}{7} ? R e q = 7 8 ?
Examples → \rightarrow → → \rightarrow → Example → \rightarrow → → \rightarrow →
Wheatstones Bridge Meter Bridge And Potentiometer L-10 Wheatstone's Bridge
Measure resistance in range 1 Ω 1 \Omega 1Ω 10 6 Ω 10^6 \Omega 1 0 6 Ω
R 1 , R 2 R_1, R_2 R 1 , R 2 R 3 R_3 R 3 R 4 R_4 R 4
Null Deflection resistance
Adjust the variable resistance R 3 R_3 R 3
I 1 R 1 = I 2 R 2 {I_1}{R_1} = {I_2}{R_2} I 1 R 1 = I 2 R 2
I 1 R 3 = I 2 R − x {I_1}{R_3} = {I_2}{R-x} I 1 R 3 = I 2 R − x
R 1 R 3 = R 2 R x ⇒ R x = R 2 R 1 ⋅ R 3 \frac{R_1}{R_3}=\frac{R_2}{R_x} \Rightarrow R_x=\frac{R_2}{R_1} \cdot R_3 R 3 R 1 = R x R 2 ⇒ R x = R 1 R 2 ⋅ R 3
Example → \rightarrow → → \rightarrow → Wheatstone's Bridge → \rightarrow → → \rightarrow →
Wheatstones Bridge Meter Bridge And Potentiometer L-10 Example
R 1 = 6 Ω R_1=6 \Omega R 1 = 6Ω
R 2 = 1.5 R_2=1.5 R 2 = 1.5
R 3 = 8 Ω R_3=8 \Omega R 3 = 8Ω
R x = R 2 R 1 ⋅ R 3 = 2 Ω R_x=\frac{R_2}{R_1} \cdot R_3=2 \Omega R x = R 1 R 2 ⋅ R 3 = 2Ω
Suppose R x = 2.01 Ω R_x=2.01 \Omega R x = 2.01Ω
Example → \rightarrow → → \rightarrow → Example → \rightarrow → → \rightarrow →
Wheatstones Bridge Meter Bridge And Potentiometer L-10 Example
Wheatstone's Bridge → \rightarrow → → \rightarrow → Example → \rightarrow → → \rightarrow →
Wheatstones Bridge Meter Bridge And Potentiometer L-10 Wheatstone's and Meter Bridge
Is a balanced wheatstone's bridge.
No current through CD.
R e q = R R_{eq} = R R e q = R
Meter Bridge
No ourrent through Galvanometes.
R S = R A D R D B = ρ l / A ρ ( 100 − 1 ) / A \frac{R}{S}=\frac{R_{A D}}{R_{D B}}=\frac{\rho l / A}{\rho (100-1) / A} S R = R D B R A D = ρ ( 100 − 1 ) / A ρl / A
ρ \rho ρ Ω − c m \Omega-\mathrm{cm} Ω − cm
l in cm
= l 100 − l =\frac{l}{100-l} = 100 − l l
Example → \rightarrow → → \rightarrow → Wheatstone's and Meter Bridge → \rightarrow → → \rightarrow →
Wheatstones Bridge Meter Bridge And Potentiometer L-10 Null Deflection
Three way key
ε 1 ε 2 = l 1 l 2 \frac{\varepsilon_1}{\varepsilon_2}=\frac{l_1}{l_2} ε 2 ε 1 = l 2 l 1
Null Deflection
A P ε 1 r 1 13 G N 1 AP\varepsilon_{1}r_{1}13GN_{1} A P ε 1 r 1 13 G N 1
There is a potential drop in the wire A B A B A B
Potential Drop across A N 1 = P . D A N_1=P . D A N 1 = P . D
I R A N 1 = ρ L 1 A = ε 1 I R_{A N_1}=\rho \frac{L_1}{A}=\varepsilon_1 I R A N 1 = ρ A L 1 = ε 1 P ε 1 P\varepsilon_1 P ε 1
I R A N 2 I R_{AN_2} I R A N 2 ρ L 2 A \rho \frac{L_2}{A} ρ A L 2 ε 2 \varepsilon_2 ε 2
Example → \rightarrow → → \rightarrow → Null Deflection → \rightarrow → → \rightarrow →
Wheatstones Bridge Meter Bridge And Potentiometer L-10 Finding Internal Resistance of a Battery
k 1 k_1 k 1 k 2 k_2 k 2 N 1 N_1 N 1
current through A B : I = V R V + R ′ A B: I=\frac{V}{R_V+R^{\prime}} A B : I = R V + R ′ V
R ′ = R^{\prime}= R ′ = A B A B A B
Potential Gradient
Φ = V ⋅ R ′ R V + R ′ , / L \Phi=\frac{V \cdot R^{\prime}}{R_V+R'}, / L Φ = R V + R ′ V ⋅ R ′ , / L
: ε = Φ R 1 : \quad \varepsilon=\Phi R_1 : ε = Φ R 1
For Null Deflection
Wheatstone's and Meter Bridge → \rightarrow → → \rightarrow → Finding Internal Resistance of a Battery → \rightarrow → → \rightarrow →
Wheatstones Bridge Meter Bridge And Potentiometer L-10 Potential Drop
K2 closed: the cell ε \varepsilon ε
I ′ = ε R + r I^{\prime}=\frac{\varepsilon}{R+r} I ′ = R + r ε
Potential drop across C D C D C D
ε − I ′ r = Φ l 2 \varepsilon-I{ }^{\prime} r=\Phi l_2 ε − I ′ r = Φ l 2
= ϵ R R + r =\frac{\epsilon R}{R+r} = R + r ϵ R
ε = Φ l 1 \varepsilon=\Phi l_1 ε = Φ l 1
r = l 1 − l 2 l 2 ⋅ R r=\frac{l_1-l_2}{l_2} \cdot R r = l 2 l 1 − l 2 ⋅ R
Null Deflection → \rightarrow → → \rightarrow → Potential Drop → \rightarrow → → \rightarrow →
Wheatstones Bridge Meter Bridge And Potentiometer L-10 Thank You
Finding Internal Resistance of a Battery → \rightarrow → → \rightarrow → Thank You → \rightarrow → → \rightarrow →
Resume presentation
Wheatstones Bridge Meter Bridge And Potentiometer L-10 Wheatstone's Bridge, Meter Bridge and Potentiometer $\rightarrow$ $\rightarrow$ Wheatstone's Bridge, Meter Bridge and Potentiometer $\rightarrow$ Rules $\rightarrow$ Example
