(A) The magnitude of current flowing through R1 is 7.2 A.
(B) The magnitude of current flowing through R2 is 1.2 A.
(C) The magnitude of current flowing through R3 is 4.8 A.
(D) The magnitude of current flowing through R5 is 2.4 A.
From KCL
\(i _ 1 +i _ 2 +i _3 =0\)
\(⇒ \frac{18-V_0}{\frac{3}{2}} +\frac{12-V_0}{\frac{1}{2}}+\frac{0-V_0}{\frac{3}{2}}=0\)
\(⇒18−V _ 0 +36−3V _0 −V _ 0 =0\)
\(⇒54=5V _ 0 \)
\(\frac{2(\frac{54}{5}-v')}{1}+\frac{18-v'}{1}=0\)
\(⇒v ′ =\frac{198}{ 5×3} = \frac{ 66}{ 5 } V\)
\(I _{ R 1} = \frac{36}{5} =7.2A\)
\(I _{ R 2} = \frac{6}{5} =1.2A\)
\(I _{ R 3} = \frac{24}{5} =4.8A\)
\(I _ {R 5} = \frac{12}{5} =2.4A\)
Find output voltage in the given circuit.
A | B | Y |
0 | 0 | 1 |
0 | 1 | 0 |
1 | 0 | 1 |
1 | 1 | 0 |
Let $ S $ denote the locus of the point of intersection of the pair of lines $$ 4x - 3y = 12\alpha,\quad 4\alpha x + 3\alpha y = 12, $$ where $ \alpha $ varies over the set of non-zero real numbers. Let $ T $ be the tangent to $ S $ passing through the points $ (p, 0) $ and $ (0, q) $, $ q > 0 $, and parallel to the line $ 4x - \frac{3}{\sqrt{2}} y = 0 $.
Then the value of $ pq $ is
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
Resistance is the measure of opposition applied by any object to the flow of electric current. A resistor is an electronic constituent that is used in the circuit with the purpose of offering that specific amount of resistance.
R=V/I
In this case,
v = Voltage across its ends
I = Current flowing through it
All materials resist current flow to some degree. They fall into one of two broad categories:
Resistance measurements are normally taken to indicate the condition of a component or a circuit.