Find $V_x$ (or $v_x(t)$) and $I_x$ (or $i_x$) using voltage division rule.
a)

b)

c)

d)

Solution

a)

Voltage divider: $V_x=\frac{5\Omega}{2\Omega+5\Omega}\times 14 V=10 V$
Ohm's law: $I_x=\frac{V_x}{5 \Omega}=2 A$

b)

Voltage divider: $V_x=\frac{4\Omega}{2\Omega+3\Omega+1\Omega+4\Omega}\times (-9 V)=-3.6 V$
Ohm's law: $I_x=-\frac{V_x}{4 \Omega}=0.9 A$
Please note that $I_x$ is leaving from the positive terminal of $V_x$. Therefore, applying the Ohm's law results in $V_x=-R\times I_x$.

c)

Voltage divider: $v_x(t)=\frac{5\Omega}{2\Omega+5\Omega+3\Omega}\times (-5 \sin (2t))=-2.5 \sin (2t) V$
Ohm's law: $i_x(t)=\frac{v_x(t)}{5 \Omega}=-\frac{1}{2} \sin (2t) A$

d)

The tricky part in this problem is the polarity of $V_x$. In the defined formula for voltage divider, the current is leaving the voltage source from the positive terminal and entering to resistors from positive terminals. In this problem, the current is entering to the the resistor from the negative terminal. Therefore, the voltage for $V_x$ is the negative of the voltage obtained from the voltage divider formula. The reason is that another voltage can be defined with the inverse polarity and its value can be found using the voltage division rule. $V_x$ is the negative of the defined voltage because it represents the voltage across the same nodes with inverse polarity.
Voltage divider: $V_x=- \frac{5\Omega}{2\Omega+5\Omega}\times 10 V= - 7.143 V$

Ohm's law ($I_x$ is entering from the negative terminal of $V_x$): $I_x= - \frac{V_x}{5 \Omega}=1.428 A$

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1. mekuria says:

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1. Done. Thanks

2. Magomu desmond says:

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