By Syed Nasar

Schaum’s strong problem-solver offers 3,000 difficulties in electrical circuits, absolutely solved step by step! The originator of the solved-problem advisor, and scholars’ favourite with over 30 million research publications bought, Schaum’s deals a diagram-packed timesaver that will help you grasp all kinds of challenge you’ll face on checks.

difficulties disguise each quarter of electrical circuits, from uncomplicated devices to complicated multi-phase circuits, two-port networks, and using Laplace transforms. move on to the solutions and diagrams you would like with our distinct, cross-referenced index. appropriate with any school room textual content, Schaum’s 3000 Solved difficulties in electrical Circuits is so entire it’s the suitable software for graduate or specialist examination prep!

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**Extra resources for 3000 Solved Problems in Electrical Circuits**

**Sample text**

I From Prob. 84 For the circuit of Fig. 3-27a determine 14 , Thus calculate the power loss in each resistor. Verify that the sum of the power losses is the same as the power delivered by the source. ~;9 A I LP lOSS V= ! 85 15 = 16 = = 120 x 40 = 4800 W Figure 3-28a shows a ladder network. 022% From the data, determine the voltage Vx and the current Ix' I The circuit reduction is shown in Fig. 3-28b thro~gh e. P From Fig. ::..... =2A 4+2 From Fig. 3-28d: 6 I = ---- 2 = 1 333 A From Fig. 3-28b: Ix = 3!

Finally, R l _ 3 = 8 O. 110 Find the total resistance between points 1 and 2 shown in Fig. 3-40a. Subsequent 42 D CHAPTER 3 2. 11. 10 (I.. ,/\/'of... L2. 1<-. fL. 6,,{l.... (l. ,HIP ~7~ =-7. b4SL

I Solving for 12 from the mesh equations of Prob. 38 12 = 5 A. Thus, P30 v = 30 x 5 = 150 w. How much current flows through the 1-n resistor of the network of Fig. 4-21a? I From Fig. l = 10 - 5 x 2 = 0 or Find the value of R in Fig. 4-22 such that the power supplied by the 100-V source to the network is the same as the power supplied by the 5-A source. Fig. :::: __ V-lOO R 10 I At node 1: V-lOO ------w-+1 At node 2: _ 100 _ 5 4 - For equal power: (1) 20 - (2) LOCI/ I = 5 V (3) From Eqs. (1) and (2) we obtain: \' V R +10 = 15 (4) V 14 +10 = 15 (5) V Thus, R J.