Question

3. Determine the equivalent resistance of the circuit. Then determine the current in each path, and the voltage across each resistor. Then determine the power dissipated by each resistor and the total power dissipated by the circuit. 10.0 12 W 4.000 5.00 2.00 3.002 8.00 V HE

Answer 1

The circuit is redrawn below.

Rs=102 R4=452 DE w L K H R1 =22 4 WM RE52 M B N C E M R5=32 N = 8V HE 0 of

• Here the resistance R3 and R4 are in parallel combination. Let their effective resistance be Ra. It s given by,

$R_{a}=\frac{R_{2}R_{3}}{R_{2}+R_{3}}=\frac{5\times 10}{5+10}=3.33\Omega$

Now Ra and R4 are in series. Let their effective resistance be Rb. It s given by,

R = R + R4 = 3.33 +4= 7.3322

Now Rb and R5 are in parallel combination. Let their effective resistance be Rc. It s given by,

7.33 X 3 RC= RoR5 Rb + R5 = 2.1322 7.33 + 3

Now Rc and R1 are in series. Their effective resistance is the equivalent resistance of the circuit and is given by,

Req Rc + R1 = 2.13 + 2 = 4.1312

So the equivalent resistance of the circuit is 4.13$\Omega$.

• The total current through the circuit is,

$I = \frac{V}{R_{eq}}=\frac{8}{4.13}=1.94A$

From the figure, it is clear that the current through R1 is,

$I_{1}=I=1.94A$

At node C, the current splits into two parts. By current division formula the current through R5 is given by,

$I_{5}=\frac{IR_{b}}{R_{b}+R_{5}}=\frac{1.94\times 7.33}{7.33+3}=1.38A$

Now the current through Rb is,

$I_{b}=I-I_{5}=1.94-1.38=0.56A$

From the figure, it is clear that the current through Rb is the current through R4. Therefore,

But at node F, the 0.56A current splits inti two parts. By current division formula the current through R2 is given by,

$I_{2}=\frac{I_{b}R_{3}}{R_{2}+R_{3}}=\frac{0.56\times 10}{10+5}=0.37A$

Now the current through R3 is,

$I_{3}=I_{b}-I_{2}=0.56-0.37=0.19A$

• The voltage across R1 is,

$V_{1}=I_{1}R_{1}=1.94\times 2=3.88V$

The voltage across R2 is,

$V_{2}=I_{2}R_{2}=0.37\times 5=1.85V$

The voltage across R3 is,

$V_{3}=I_{3}R_{3}=0.19\times 10=1.9V$

The voltage across R4 is,

$V_{4}=I_{4}R_{4}=0.56\times 4=2.24V$

The voltage across R5 is,

$V_{5}=I_{5}R_{5}=1.38\times 3=4.14V$

• The power dissipated by R1 is,

$P_{1}=I{_{1}}^{2}R_{1}=(1.94)^{2}\times 2=7.53W$

The power dissipated by R2 is,

$P_{2}=I{_{2}}^{2}R_{2}=(0.37)^{2}\times 5=0.68W$

The power dissipated by R3 is,

$P_{3}=I{_{3}}^{2}R_{3}=(0.19)^{2}\times 10=0.36W$

The power dissipated by R4 is,

$P_{4}=I{_{4}}^{2}R_{4}=(0.56)^{2}\times 4=1.25W$

The power dissipated by R5 is,

$P_{5}=I{_{5}}^{2}R_{5}=(1.38)^{2}\times 3=5.71W$

• The total power dissipated by the circuit is,

$P=VI=8\times 1.94=15.52W$

So the total power dissipated by the circuit is 15.52W.