in the given circuit find the potential of point E .pls show how to solve
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Hello Dear.
Given ⇒
e.m.f.( =ε) = 8 V.
R₁ = 5 Ω
R₂ = 1 Ω
From the Figure we can observe that the Resistors are in series.
∴ R = R₁ + R₂
= 5 + 1
= 6 Ω
Using Ohm's law,
V = I × R
⇒ I = V/R
⇒ I = 8/6
∴ I = 4/3 A.
∵ Current remains same in the Series,
∴ Current flowing through the 1 Ω Resistors is 4/3 A.
Again Using the Ohm's law,
V = I × R
V = 4/3 × 1
∴ V = 4/3 V.
∴ Potential across the Point E is 1.33 A.
Hope it helps.
Given ⇒
e.m.f.( =ε) = 8 V.
R₁ = 5 Ω
R₂ = 1 Ω
From the Figure we can observe that the Resistors are in series.
∴ R = R₁ + R₂
= 5 + 1
= 6 Ω
Using Ohm's law,
V = I × R
⇒ I = V/R
⇒ I = 8/6
∴ I = 4/3 A.
∵ Current remains same in the Series,
∴ Current flowing through the 1 Ω Resistors is 4/3 A.
Again Using the Ohm's law,
V = I × R
V = 4/3 × 1
∴ V = 4/3 V.
∴ Potential across the Point E is 1.33 A.
Hope it helps.
Anonymous:
It is right bro, Potential across the Resistors of 1 ohm, same for the resistor of Point E. Because you must know that when current flows in the conductor It reaches the Point E in the last. Before Point E it reaches the 1 ohm Resistors where Potential is decreases now the potential will be same for the point E.
Because the Potential which was decreases in 1 ohm resistors now will not decrease as there is not any resistors after that, thus the Potential at point E will be the same as the potential in Resistor of 1 Ohm.
Isn't the potential at the negative terminal of the circuit always zero?
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