Problem 1: Figure shows three resistors, R1, R2 and R3 connected across each other, i.e. in parallel, across a battery source
of V volts.

**PARALLEL NETWORKS:**

**Problem 1: **Figure shows three resistors, R1, R2 and R3 connected across each other, i.e. in** **parallel, across a battery source

of V volts.

**In a parallel circuit:**

(a) the sum of the currents I1, I2 and I3 is equal to the total circuit current, I, i.e. I =I1 +I2 +I3, and

the source p.d., V volts, is the same across each of the

From Ohm’s law:

I1 = V/R1

, I2 = V/R2

, I3 = V/R3 and I = V/R

where R is the total circuit resistance. Since I =I1 + I2 + I3

then

V/R= V/R1+ V/R2+ V/R3 Dividing throughout by V gives:

This equation must be used when finding the total resistance R of a parallel circuit. For the special case of two resistors in parallel

**Problem 2: **For the circuit shown in Figure , determine (a) the reading on the ammeter, and (b) the** **value of resistor R2.

P.d. across R1 is the same as the supply voltage V.

Hence supply voltage, V =8 × 5=40V

(a) Reading on ammeter, I = V R3= 40/20=2A

Current flowing through R2 =11−8−2=1A

Hence, R2 = V/I2= 40/1=40 Ω

(a) The total circuit resistance R is given by 1/R= 1/R1+ 1/R2= 1/3+ 1/6

1/R= 2 + 1/6= 3/6 Hence, R = 6/3= 2 Ω

(b) Current in the 3 Ωresistance, I1 = V R1= 12/3= 4A

**Problem 3: **For the circuit shown in Figure find (a) the value of the supply voltage V and (b) the** **value of current I.

(a) P.d. across 20 Ωresistor = I2R2 = 3× 20 = 60V, hence supply voltage V =60V since the circuit is connected in parallel.

(b)Current I1 = V/R1= 60/10= 6A; I2 = 3A

I3 = V/R3= 60/60= 1A

Current I =I1+I2+I3 and hence I =6+3+1=10A Alternatively,

1/R= 1/60+ 1/20+ 1/10= 1 + 3 + 6/60= 10/60 Hence total resistance R= 6010=6 Ω Current I = V/R= 60/6=10A

**Problem 4: **Find the equivalent resistance for the circuit shown in Figure

*R*3,* R*4 and*R*5 are connected in parallel and their equivalent resistance* R *is given by:* *1/*R*= 1/3+ 1/6+ 1/18=6 + 3 + 1/18= 10/18

Hence *R*= 18/10=1*.*8 Ω

The circuit is now equivalent to four resistors in series and the equivalent circuit resistance =1+2*.*2+1*.*8+4=9 Ω

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