I need answers of question 2 to question 5 Circuit with Two Batteries and Six Re
ID: 3280444 • Letter: I
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I need answers of question 2 to question 5
Circuit with Two Batteries and Six Resistors A circuit is constructed with six resistors and two batteries as shown. The battery voltages are V1 -18 V and V2 12 V. The positive terminals are indicated with a sign, The values for the resistors are:w-. R1-R5-53 , R2-R6-102 R3-42 , and R4-112 . The positive directions for the currents l1, 12 and 13 are indicated by the directions 3 of the arrows. a Rb R, 1) What is V4, the magnitude of the voltage across the resistor R4? 8.145 V Submit Your submissions: 8.145 Computed value: 8.145Submitted: Tuesday, October 10 at 8:35 AM Feedback: Your answer has been judged correct; the exact answer is: 8.14545454545454.Explanation / Answer
1) By kirchoff's law, we have V2 - i4* ( R5 + R4) =0
12 - i4 * ( 53+112) = 0
i4 = 0.07273 Amperes
magnitude of the voltage across the resistor R4 = V4 = i4 * R4
= 0.0273* 112
= 8.145 Volts
2) By kirchoff's law, we have
V2 - i2*R2 - i1*R1 - i2*R6 = 0
12- (i2* 102) - (i1*53) - (i2*102) =0
12 - (i1* 53 ) - (i2*204)= 0 ----> (1)
By kirchoff's law, we have
V2 - i2*R2 - i3*R3 -V1 - i2*R6 =0
12 - (i2*102) - (i3*42) - 18 - ( i2*102) = 0
-6 - (i2*204) - (i3*42) = 0
we have i3 = i2 - i1 , substituting in the above expression
-6 - (i2*204) - ((i2-i1)*42) =0
-6 - (i2*246) + (i1*42) =0 ----> (2)
Solving equations 1 , 2
we get i1 = 0.1933 Amperes.
i2 = 0.0086087 Amperes.
i3 = i2 -i1 = 0.0086087 - 0.1933 = -0.185 Amperes = 0.185 A
3) i2 = 0.0086087 Amperes.
4) i1 = 0.1933 Amperes.
5) V(a) - V(b) = i2*R6 = 0.0086087 * 102
= 0.88 Volts.
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