WebAt t = 0-, u(t) = 0. Consider the circuit shown in Figure (a). i L(0-) = 0, and v R(0-) = 0. But, -v R(0-) + v C(0-) + 10 = 0, or v C(0-) = -10V. (a) At t = 0+, since the inductor current and … http://www.phys.ufl.edu/~chungwei/phy2054_fall_2011/sol/q05-4350sol.pdf
For the circuit shown in the figure the current through the
WebDoing DC analysis of the transistor circuit is the most common way of finding out the value of I B in the circuit. The equation to solve for I e is: So we must solve for V bb and R B in … WebAs Kirchhoff’s junction rule states that : I1 = I2 + I3. The supply current flowing through resistor R1 is given as : 1.0 + 0.5 = 1.5 Amps. Thus I1 = IT = 1.5 Amps, I2 = 1.0 Amps … emhr grant workshop
Solved 36) 36) See Figure 6.3, What is the current drawn - Chegg
Web(a) At t = 0-, the equivalent circuit is shown in Figure (a). 60 20 = 15 kohms, [latex]i_{R}(0-)[/latex] = 80/(25 + 15) = 2mA. By the current division WebAnswer (1 of 20): If the 3ohm resistor is passing 800mA, then so must the other one as they’re connected in parallel. So the two resistors together are passing 1600mA This means that the 4ohm resistor must also be passing 1600mA Ohms law tells us V = I x R which would give us 1.6 x 4 which is 6... WebAs Kirchhoff’s junction rule states that : I1 = I2 + I3. The supply current flowing through resistor R1 is given as : 1.0 + 0.5 = 1.5 Amps. Thus I1 = IT = 1.5 Amps, I2 = 1.0 Amps and I3 = 0.5 Amps and from that information we could calculate the I*R voltage drops across the devices and at the various points (nodes) around the circuit. dprly