Chapter 28, Problem 1CQ

Rank in order, from largest to smallest, the currents $I_a$to $I_d$ through the four resistors in FIGURE Q28.1.

To determine

The rank in order of currents from largest to smallest

Answer to Problem 1CQ

Solution:

The rank in order of currents from largest to smallest is $I_c>I_a=I_d>I_b$

Explanation of Solution

Given Info:

Voltage = $V_a=2.0\Omega$

Resistance= $R_a=2.0\Omega$

Current = $I_a=?$

Voltage = $V_b=1.0\Omega$

Resistance= $R_b=2.0\Omega$

Current = $I_b=?$

Voltage = $V_c=2.0\Omega$

Resistance= $R_c=1.0\Omega$

Current = $I_c=?$

Voltage = $V_d=1.0\Omega$

Resistance= $R_d=1.0\Omega$

Current = $I_d=?$

Formula:

Ohm’s law => $I=\frac{V}{R}$ ------ (1)

Applying eqn(1) to all the sets, we get

$I_a=\frac{V_a}{R_a}$ -------- (2)

$I_b=\frac{V_b}{R_b}$ -------- (3)

$I_c=\frac{V_c}{R_c}$ -------- (4)

$I_d=\frac{V_d}{R_d}$ -------- (5)

Calculation:

Substituting the given data in eqn(2)

⇒ $I_a$ = 1 A

Substituting the given data in eqn(3)

⇒ $I_b$ = 0.5 A

Substituting the given data in eqn(4)

⇒ $I_c$ = 2 A

Substituting the given data in eqn(5)

⇒ $I_d$ = 1 A

On comparing the above values, we get the required relation is $I_c>I_a=I_d>I_b$

Conclusion:

The rank in order of currents from largest to smallest is $I_c>I_a=I_d>I_b$