In circuit theory, ThÃ©venin's theorem for linear electrical networks states that any combination of voltage sources, current sources, and resistors with two terminals is electrically equivalent to a single voltage source V and a single series resistor R. For single frequency AC systems the theorem can also be applied to general impedances, not just resistors.

**THEVENINS THEOREM**:

In circuit theory, **ThÃ©venin's theorem** for linear electrical networks states that any combination of voltage sources, current sources, and resistors with two terminals is electrically equivalent to a single voltage source V and a single series resistor R. For single frequency AC systems the theorem can also be applied to general impedances, not just resistors.

The procedure adopted when using ThÃ©veni determine the current in any branch of an active network (i.e. one containing a source of e.m.f.):

(i) remove the resistance R from that branch,

(ii) determine the open-circuit voltage, E, across the break,

remove each source of e.m.f. and replace them by their internal resistances and then determine the resistance, r, â€˜looking-inâ€™break.

(iv) determine the value of the current from the equivalent circuit shown in Figure 13.33, i.e. **I = ****ER+r**

**Problem 1:** Use ThÃ©veninâ€™s theorem to find the current flowing in the 10 Î© resistor for the circuit** **shown in Figure

Following the above procedure:

The 10 Î©resistance is removed from the circuit as shown in Figure There is no current flowing in the 5 Î©resistor and current I1 is given by:

I1 = 10/R1 + R2

= 10/2 + 8

= 1A

P.d. across R2 =I1R2 =1Ã—8=8V Hence p.d. across AB, i.e. the open-circuit voltage across the break, E =8V

(iii) Removing the source of e.m.f. gives the circuit of Figure Resistance, r = R3 + R1R2/R1 + R2

=5+ (2Ã—8/2+8)

= 5 + 1.6 = 6.6 Î©

(iv) The equivalent ThÃ©veninâ€™s circuit is shown in F

Current I = E/R+r= 8/10+6.6= 8/16.6=0.482A

**Problem 2: **For the network shown in Figure determine the current in the 0.8** **Î©** **resistor using

ThÃ©veninâ€™s theorem.

Following the procedure:

The 0.8_ resistor is removed from the circuit as shown in Figure

Current I1 = 12/1+5+4= 12/10

=1.2A

P.d. across 4 Î©resistor=4I1 =(4) (1.2)=4.8V

Hence p.d. across AB, i.e. the open-circuit voltage across AB, E =4.8V

(iii) Removing the source of e.m.f. gives the circuit shown in Figure (c). The equivalent circuit of Figure

(c) is shown in Figure (d), from which,resistance r = 4Ã—6/4+6= 24/10

=2.4 Î©

(iv) The equivalent ThÃ©veninâ€™s circuit I=E/+R is show

= 4.8/2.4+0.8

= 4.8/3.2

I = 1.5A=current in the 0.8 Î©resistor

**Problem 3: **U Use ThÃ©veninâ€™s theorem to determine the current I flowing in the 4 Î© resistor shown in Figure.

Find also the power dissipated in the 4 Î©resistor.

(i) The 4 Î©resistor is removed from the circuit as shown in Figure

(ii) Current I1 = E1 âˆ’E2/r1 +r2

= 4âˆ’2/2+1

= 2/3A

(iii) Removing the sources of e.m.f. gives the circuit shown in Figure (c), from which resistance

r = 2 Ã— 1/2 + 1 = 2/3 Î©

(iv)The equivalent ThÃ©veninâ€™s circuit is show

= 8/14

= 0.571A

= current in the 4 Î©resistor

**Problem 4: **Power dissipated in 4** **Î©resistor,P=I2R=(0.571)2 (4)=1.304W** **Use ThÃ©veninâ€™s t** **to determine the current flowing in the 3 Î© resistance of the network shown in Figure (a). The

voltage source has negligible internal resistance.

i) The 3 Î©resistance is removed from the circuit as shown in Figure (b).

(ii) The 1 2/3 Î©resistance now carries no current. P.d. across 10 Î©resistor=(10/10+5)(24)

=**16V**

Hence p.d. across AB, E =16V

(iii) Removing the source of e.m.f. and replacing it by its internal resistance means that the 20 Î©resistance is short-circuited as shown in Figure (c) since its internal resistance is zero. The 20 Î©resistance may thus be removed as shown in Figure (d)

From Figure (d), resistance,

r =1 2/3+ 10Ã—5/10+5 =5 Î©

(iv) The equivalent ThÃ©veninâ€™s circuit is show current, I = E/r +R= 16/3+5= 16/8

=2A

**Problem 5: **AWheatstone Bridge network is shown in Figure (a). Calculate the current flowing in the 32 Î©resistor, and its direction, using ThÃ©veni negligible resistance.

The 32 Î©resistor is removed from the circuit as shown in Figure (b) The p.d. between A and C,

VAC =R1/R1 +R4 (E) =2/2+11(54) = 8.31V The p.d. between B and C,

VBC =R2/R2 +R3 (E) =14/14+3(54) = 44.47V Hence the p.d. between A and B=44.47âˆ’8.31=36.16V

Point C is at a potential of +54V. Between C and A is a voltage drop of 8.31V. Hence the voltage at point A is 54âˆ’8.31=45.69V. Between C and B is a voltage drop of 44.47V. Hence the voltage at point B is 54âˆ’44.47=9.53V. Since the voltage at A is greater than at B, current must flow in the direction A to B.

Replacing the source of e.m.f. with a short-circuit (i.e. zero internal resistance) gives the circuit shown in Figure (c). The circuit is redrawn and simplified as shown in Figure (d) and (e), from which the resistance between terminals A and B,

r = 2 Ã— 11/2 + 11+ 14 Ã— 3/14 + 3

= 22/13+ 42/17

= 1.692 + 2.471 = 4.163 Î©

(iv) The equivalent ThÃ©veninâ€™s circuit is show

current I = E/r +R5

= 36.16/4.163+32 =1A

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Electrical and electronics : Circuit Theory : Network Reduction and Network Theorems for DC And AC Circuits : Thevenins Theorem |

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