A resistor of 6 ohms is connected in a series with another resistor of 4 ohms. A potential difference of 20 V is applied across the combination. What is the current in the circuit and the potential difference across the 6 ohms resistor?
In series, the total resistance is 10 ohms. Using Ohm’s law, V = IR, I = V/R, and I = 20/10 = 2AThe 20V is divided among the resistors in an inverse proportion, with the lower resistance dissipating more power and therefore more voltage (since the current through both resistors is the same). But since we already know how much current is passing through the 6 ohm resistor (2A), we can just use Ohm’s law again.V = IR … V=2*6 = 12V
In a series RL circuit contains two resistors and two inductors. The resistors are 27 ohms and 47 ohms.?
Series circuit the current is the same through all ckt. elements. X(total) = 50+40 = 90 ohm, R=27+47 = 74 ohm The impedance: Z = √ 74² +90² = 116.5 ohm The series current: I = V/Z = 120/116.5 = 1.03 A The voltage across the 40 ohm reactance is: V = IX = 1.03(40) = 41.2 volts
When two resistors are connected in parallel, their equivalent resistance is 2, and when they are connected in series, their equivalent resistance is 9. What are the values of the resistors?
From the Given question, I can easily depict that:WHEN RESISTORS CONNECTED IN SERIES:R1 + R2 = 9 ------------------------eq(1)WHEN RESISTORS CONNECTED IN PARALLEL:1/R1 + 1/R2 = 2OR(R1R2)/(R1+R2) = 2 ---------------------eq(2)Please see the attached photo for further explanation.Thank you.
If R is the total resistance for a parallel circuit with two resistors of resistances r1 and r2 , .?
1/r1 = 1/R - 1/r2 1/r1 = 1/20 - 1/75 1/r1 = (75 - 20)/20*75 r1 = 1500/55 r1 = 27 ohms
In a circuit consisting of a 2.2K ohm resistor and a 5.6K ohm resistor in parallel, in series with another par?
The equivalent resistance of two resistors in parallel R1||R2 = R1*R2/(R1+R2) 2.2k||5.6k = 1579.5Ω 1k||470Ω = 319.7Ω So the total resistance Req = 1,579.5Ω+319.7Ω+10,000Ω = 11,899.2Ω The current in the circuit is 50V/11,899.2Ω = 4.202ma The voltage across the 10k resistor = 4.202e-3A*10e3Ω = 42.02V <--------------- Ans
In a electric circuit two resistors of 2ohm and 4 ohm respectively are connected in series to a 6V battery.?
Hello, I = V/R = 6/(4+2) = 1 Amp. P = (I^2) * R = (1^2) * 4 = 4 W Energy = Power * Time = 4 * 5sec = 20 Ws... = 20 J............
If two resistors of resistances 3 ohm and 2 ohm are connected in parallel to a 6 volt battery, then what is the total current in the circuit?
[math]R equivalent = \dfrac{2\times 3}{2 + 3} = \dfrac{6}{5} = 1 .2 \,\Omega[/math][math]I\, total = \dfrac{6}{1.2} = 5 A[/math]
A circuit consisting of a resistance of 12 ohms and an inductance of 0,2 Henry in series is connected to a 250V, 50Hz supply. What current will flow in the circuit?
First, calculate the inductive reactance of 0.2 H at 50 Hz, which is given by 2piFL and expressed in ohms:XL = 2piFL = 2 x 3.1416 x 50 Hz x .2 H = 62.83 ohmsThen add the above value to the series resistance of 12 ohms:62.83 ohms + 12 ohms = 74.83 ohmsUsing Ohm's’ Law, calculate the current through 74.8 ohms when applying a potential of 250 volts:E/R = I250 volts / 74.83 ohms = 3.34 amperes