48. The circuit of Figure P10.48 is a general Wheatstone bridge circuit (the dc

Question

48. The circuit of Figure P10.48 is a general Wheatstone bridge circuit (the dc version of which is described in Problem 35 of Chapter 2). Here the circuit is used to measure the value of the unknown inductance L. (a) Suppose R, = 0. Show that the steady-state voltage vout(t) = 0 when RcRL = L/C. Note: In general the condition for a null voltage, v(t) = 0, in the steady state is that the products of the cross impedances be equal. (b) Again suppose R5 = 0. You are given that RcC = 2 sec and that the voltage source vin (t) is a sinusoid with a frequency of 5 rad/sec. With the unknown inductance L inserted in the circuit as shown, you adjust RL until you reach a sinusoidal steady-state voltage null, vout(t) = 0 V. The resulting value for RL is 3 Ohm. Find the value of L. Now suppose R5 not equal to 0. Show that the condition of part (a) is still valid. You can do this with some straightforward reasoning without writing any equations.

Explanation / Answer

a) => Zc / Rc = RL / ZL

=> Rc * RL = Zc * ZL

=> Rc * RL = (1 / jwC) * (jwL) = jwL / jwC = L / C

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b) given w = 5 rad/s

=> Rc RL = L / C

so L =  (Rc*C) RL

=> L = 2 RL = 2*3 = 6 H

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c) changing Rs won't change the fact that :

Ic Zc = IL RL

and

Ic Rc = IL ZL

so dividing these eqns gives :

Zc/Rc = RL / ZL

so Rc * RL = Zc * ZL = L / C

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