Question

(6) $i = 0.1\\,\text{a}, r = 560\\,\omega$

1. calculate the voltage for each value of $i$ and $r$.

(a) $i = 1\\,\text{ma}, r = 10\\,\omega$
(b) $i = 50\\,\text{ma}, r = 33\\,\omega$
(c) $i = 3\\,\text{a}, r = 5.6\\,\text{k}\omega$
(d) $i = 1.6\\,\text{ma}, r = 2.2\\,\text{k}\omega$
(e) $i = 250\\,\mu\text{a}, r = 10\\,\text{k}\omega$
(f) $i = 500\\,\text{ma}, r = 1.5\\,\text{m}\omega$
(g) $i = 850\\,\text{ma}, r = 10\\,\text{m}\omega$
(h) $i = 75\\,\mu\text{a}, r = 47\\,\omega$

Explanation:

To solve for the voltage \( V \) in each case, we use Ohm's Law, which states that \( V = I \times R \). We need to ensure that the units of current \( I \) (in amperes, A) and resistance \( R \) (in ohms, \( \Omega \)) are consistent. Recall that:

• \( 1 \, \text{mA} = 10^{-3} \, \text{A} \)
• \( 1 \, \mu\text{A} = 10^{-6} \, \text{A} \)
• \( 1 \, \text{k}\Omega = 10^{3} \, \Omega \)
• \( 1 \, \text{M}\Omega = 10^{6} \, \Omega \)

Part (a): \( I = 1 \, \text{mA}, \, R = 10 \, \Omega \)

Step 1: Convert current to amperes

\( I = 1 \, \text{mA} = 1 \times 10^{-3} \, \text{A} \)

Step 2: Apply Ohm’s Law

\( V = I \times R = (1 \times 10^{-3} \, \text{A}) \times (10 \, \Omega) = 0.01 \, \text{V} \)

Part (b): \( I = 50 \, \text{mA}, \, R = 33 \, \Omega \)

Step 1: Convert current to amperes

\( I = 50 \, \text{mA} = 50 \times 10^{-3} \, \text{A} = 0.05 \, \text{A} \)

Step 2: Apply Ohm’s Law

\( V = I \times R = (0.05 \, \text{A}) \times (33 \, \Omega) = 1.65 \, \text{V} \)

Part (c): \( I = 3 \, \text{A}, \, R = 5.6 \, \text{k}\Omega \)

Step 1: Convert resistance to ohms

\( R = 5.6 \, \text{k}\Omega = 5.6 \times 10^{3} \, \Omega \)

Step 2: Apply Ohm’s Law

\( V = I \times R = (3 \, \text{A}) \times (5.6 \times 10^{3} \, \Omega) = 16800 \, \text{V} = 16.8 \, \text{kV} \)

Part (d): \( I = 1.6 \, \text{mA}, \, R = 2.2 \, \text{k}\Omega \)

Step 1: Convert units

\( I = 1.6 \, \text{mA} = 1.6 \times 10^{-3} \, \text{A} \)
\( R = 2.2 \, \text{k}\Omega = 2.2 \times 10^{3} \, \Omega \)

Step 2: Apply Ohm’s Law

\( V = I \times R = (1.6 \times 10^{-3} \, \text{A}) \times (2.2 \times 10^{3} \, \Omega) = 3.52 \, \text{V} \)

Part (e): \( I = 250 \, \mu\text{A}, \, R = 10 \, \text{k}\Omega \)

Step 1: Convert units

\( I = 250 \, \mu\text{A} = 250 \times 10^{-6} \, \text{A} = 2.5 \times 10^{-4} \, \text{A} \)
\( R = 10 \, \text{k}\Omega = 10 \times 10^{3} \, \Omega = 10^{4} \, \Omega \)

Step 2: Apply Ohm’s Law

\( V = I \times R = (2.5 \times 10^{-4} \, \text{A}) \times (10^{4} \, \Omega) = 2.5 \, \text{V} \)

Part (f): \( I = 500 \, \text{mA}, \, R = 1.5 \, \text{M}\Omega \)

Step 1: Convert units

\( I = 500 \, \text{mA} = 500 \times 10^{-3} \, \text{A} = 0.5 \, \text{A} \)
\( R = 1.5 \, \text{M}\Omega = 1.5 \times 10^{6} \, \Omega \)

Step 2: Apply Ohm’s Law

\( V = I \times R = (0.5 \, \text{A}) \times (1.5 \times 10^{6} \, \Omega) = 750000 \, \text{V} = 750 \, \text{kV} \)

Part (g): \( I = 850 \, \mu\text{A}, \, R = 10 \, \text{M}\Omega \)

Step 1: Convert units

\( I = 850 \, \mu\text{A} = 850 \times 10^{-6} \, \text{A} = 8.5 \times 10^{-4} \, \text{A} \)
\( R = 10 \, \text{M}\Omega = 10 \times 10^{6} \, \Omega = 10^{7} \, \Omega \)

Step 2: Apply Ohm’s Law

\( V = I \times R = (8.5 \times 10^{-4} \, \text{A}) \times (10^{7} \, \Omega) = 8500 \, \text{V} = 8.5 \, \text{kV} \)

Part (h): \( I = 75 \, \mu\text{A}, \, R = 47 \, \Omega \)

Step 1: Convert current to amperes

\( I = 75 \, \mu\text{A} = 75 \times 10^{-6} \, \text{A} = 7.5 \times 10^{-5} \, \text{A} \)

Step 2: Apply Ohm’s Law

\( V = I \times R = (7.5 \times 10^{-5} \, \text{A}) \times (47 \, \Omega) = 3.525 \times 10^{-3} \, \text{V} = 3.525 \, \text{mV} \)

Final Answers:

(a) \( \boldsymbol{0.01 \, \text{V}} \)
(b) \( \boldsymbol{1.65 \, \text{V}} \)
(c) \( \boldsymbol{16.8 \, \text{kV}} \)
(d) \( \boldsymbol{3.52 \, \text{V}} \)
(e) \( \boldsymbol{2.5 \, \text{V}} \)
(f) \( \boldsymbol{750 \, \text{kV}} \)
(g) \( \boldsymbol{8.5 \, \text{kV}} \)
(h) \( \boldsymbol{3.525 \, \text{mV}} \)

Answer:

To solve for the voltage \( V \) in each case, we use Ohm's Law, which states that \( V = I \times R \). We need to ensure that the units of current \( I \) (in amperes, A) and resistance \( R \) (in ohms, \( \Omega \)) are consistent. Recall that:

• \( 1 \, \text{mA} = 10^{-3} \, \text{A} \)
• \( 1 \, \mu\text{A} = 10^{-6} \, \text{A} \)
• \( 1 \, \text{k}\Omega = 10^{3} \, \Omega \)
• \( 1 \, \text{M}\Omega = 10^{6} \, \Omega \)

Part (a): \( I = 1 \, \text{mA}, \, R = 10 \, \Omega \)

Step 1: Convert current to amperes

\( I = 1 \, \text{mA} = 1 \times 10^{-3} \, \text{A} \)

Step 2: Apply Ohm’s Law

\( V = I \times R = (1 \times 10^{-3} \, \text{A}) \times (10 \, \Omega) = 0.01 \, \text{V} \)

Part (b): \( I = 50 \, \text{mA}, \, R = 33 \, \Omega \)

Step 1: Convert current to amperes

\( I = 50 \, \text{mA} = 50 \times 10^{-3} \, \text{A} = 0.05 \, \text{A} \)

Step 2: Apply Ohm’s Law

\( V = I \times R = (0.05 \, \text{A}) \times (33 \, \Omega) = 1.65 \, \text{V} \)

Part (c): \( I = 3 \, \text{A}, \, R = 5.6 \, \text{k}\Omega \)

Step 1: Convert resistance to ohms

\( R = 5.6 \, \text{k}\Omega = 5.6 \times 10^{3} \, \Omega \)

Step 2: Apply Ohm’s Law

\( V = I \times R = (3 \, \text{A}) \times (5.6 \times 10^{3} \, \Omega) = 16800 \, \text{V} = 16.8 \, \text{kV} \)

Part (d): \( I = 1.6 \, \text{mA}, \, R = 2.2 \, \text{k}\Omega \)

Step 1: Convert units

\( I = 1.6 \, \text{mA} = 1.6 \times 10^{-3} \, \text{A} \)
\( R = 2.2 \, \text{k}\Omega = 2.2 \times 10^{3} \, \Omega \)

Step 2: Apply Ohm’s Law

\( V = I \times R = (1.6 \times 10^{-3} \, \text{A}) \times (2.2 \times 10^{3} \, \Omega) = 3.52 \, \text{V} \)

Part (e): \( I = 250 \, \mu\text{A}, \, R = 10 \, \text{k}\Omega \)

Step 1: Convert units

\( I = 250 \, \mu\text{A} = 250 \times 10^{-6} \, \text{A} = 2.5 \times 10^{-4} \, \text{A} \)
\( R = 10 \, \text{k}\Omega = 10 \times 10^{3} \, \Omega = 10^{4} \, \Omega \)

Step 2: Apply Ohm’s Law

\( V = I \times R = (2.5 \times 10^{-4} \, \text{A}) \times (10^{4} \, \Omega) = 2.5 \, \text{V} \)

Part (f): \( I = 500 \, \text{mA}, \, R = 1.5 \, \text{M}\Omega \)

Step 1: Convert units

\( I = 500 \, \text{mA} = 500 \times 10^{-3} \, \text{A} = 0.5 \, \text{A} \)
\( R = 1.5 \, \text{M}\Omega = 1.5 \times 10^{6} \, \Omega \)

Step 2: Apply Ohm’s Law

\( V = I \times R = (0.5 \, \text{A}) \times (1.5 \times 10^{6} \, \Omega) = 750000 \, \text{V} = 750 \, \text{kV} \)

Part (g): \( I = 850 \, \mu\text{A}, \, R = 10 \, \text{M}\Omega \)

Step 1: Convert units

\( I = 850 \, \mu\text{A} = 850 \times 10^{-6} \, \text{A} = 8.5 \times 10^{-4} \, \text{A} \)
\( R = 10 \, \text{M}\Omega = 10 \times 10^{6} \, \Omega = 10^{7} \, \Omega \)

Step 2: Apply Ohm’s Law

\( V = I \times R = (8.5 \times 10^{-4} \, \text{A}) \times (10^{7} \, \Omega) = 8500 \, \text{V} = 8.5 \, \text{kV} \)

Part (h): \( I = 75 \, \mu\text{A}, \, R = 47 \, \Omega \)

Step 1: Convert current to amperes

\( I = 75 \, \mu\text{A} = 75 \times 10^{-6} \, \text{A} = 7.5 \times 10^{-5} \, \text{A} \)

Step 2: Apply Ohm’s Law

\( V = I \times R = (7.5 \times 10^{-5} \, \text{A}) \times (47 \, \Omega) = 3.525 \times 10^{-3} \, \text{V} = 3.525 \, \text{mV} \)

Final Answers:

(a) \( \boldsymbol{0.01 \, \text{V}} \)
(b) \( \boldsymbol{1.65 \, \text{V}} \)
(c) \( \boldsymbol{16.8 \, \text{kV}} \)
(d) \( \boldsymbol{3.52 \, \text{V}} \)
(e) \( \boldsymbol{2.5 \, \text{V}} \)
(f) \( \boldsymbol{750 \, \text{kV}} \)
(g) \( \boldsymbol{8.5 \, \text{kV}} \)
(h) \( \boldsymbol{3.525 \, \text{mV}} \)