current through resistor R₆ in the circuit shown below is 2 mA, that is,I₆ = 2 mA.
(a) Use Ohm's law to find the voltage V₄ across R₆.
(b) Use Ohm's law to find the current I₅ through R₅.
(c) Use KCL to find the current I₄ through R₄.
(d) Use Ohm's law to find the voltage V₃ across R₄.
(e) Use KVL to find the voltage V₂ across R₃.
(f) Use Ohm's law to find the current I₃ through R₃.
(g) Use Ohm's law to find the current I₂ through R₂.
(h) Use KCL to find the current I₁ through R₁.
(i) Use Ohm's law to find the voltage V₁ across R₁.
(j) Use KVL to find the voltage V_s.

Question

current through resistor R<sub>6</sub> in the circuit shown below is 2 mA, that is,I<sub>6</sub> = 2 mA. (a) Use Ohm's law to find the voltage V<sub>4</sub> across R<sub>6</sub>. (b) Use Ohm's law to find the current I<sub>5</sub> through R<sub>5</sub>. (c) Use KCL to find the current I<sub>4</sub> through R<sub>4</sub>. (d) Use Ohm's law to find the voltage V<sub>3</sub> across R<sub>4</sub>. (e) Use KVL to find the voltage V<sub>2</sub> across R<sub>3</sub>. (f) Use Ohm's law to find the current I<sub>3</sub> through R<sub>3</sub>. (g) Use Ohm's law to find the current I<sub>2</sub> through R<sub>2</sub>. (h) Use KCL to find the current I<sub>1</sub> through R<sub>1</sub>. (i) Use Ohm's law to find the voltage V<sub>1</sub> across R<sub>1</sub>. (j) Use KVL to find the voltage V<sub>s</sub>.

Quizplus · Accepted Answer

The Answer is (a)

V _ { 4 } = \mathrm { R } _ { 6 } \mathrm { I } _ { 6 } = 6 \mathrm { k } \Omega imes 2 \mathrm {~mA} = 12 \mathrm {~V}

(b)

I _ { 5 } = \frac { V _ { 4 } } { R _ { 5 } } = \frac { 12 \mathrm {~V} } { 3 k \Omega } = 4 \mathrm {~mA}

(c)

\mathrm { I } _ { 4 } = \mathrm { I } _ { 5 } + \mathrm { I } _ { 6 } = 4 \mathrm {~mA} + 2 \mathrm {~mA} = 6 \mathrm {~mA}

(d)

\mathrm { V } _ { 3 } = \mathrm { R } _ { 4 } \mathrm { I } _ { 4 } = 1 \mathrm { k } \Omega imes 6 \mathrm {~mA} = 6 \mathrm {~V}

(e)

\mathrm { V } _ { 2 } = \mathrm { V } _ { 3 } + \mathrm { V } _ { 4 } = 12 \mathrm {~V} + 6 \mathrm {~V} = 18 \mathrm {~V}

(f)

I _ { 3 } = \frac { V _ { 2 } } { R _ { 3 } } = \frac { 18 \mathrm {~V} } { 6 \mathrm { k } \Omega } = 3 \mathrm {~mA}

(g)

I _ { 2 } = \frac { V _ { 2 } } { R _ { 2 } } = \frac { 18 \mathrm {~V} } { 9 k \Omega } = 2 \mathrm {~mA}

(h)

\mathrm { I } _ { 1 } = \mathrm { I } _ { 2 } + \mathrm { I } _ { 3 } + \mathrm { I } _ { 4 } = 2 \mathrm {~mA} + 3 \mathrm {~mA} + 6 \mathrm {~mA} = 11 \mathrm {~mA}

(i)

V _ { 1 } = R _ { 1 } I _ { 1 } = 1 \mathrm { k } \Omega imes 11 \mathrm {~mA} = 11 \mathrm {~V}

(j)

\mathrm { V } _ { \mathrm { s } } = \mathrm { V } _ { 1 } + \mathrm { V } _ { 2 } = 11 \mathrm {~V} + 18 \mathrm {~V} = 29 \mathrm {~V}

Current Through Resistor R6 in the Circuit Shown Below Is

Current Through Resistor R₆ in the Circuit Shown Below Is