Question

1. determine the current in each case: (a) ( v = 5 , \text{v}, r = 1.0 , omega ) (b) ( v = 15 , \text{v}, r = 10 , omega ) (c) ( v = 50 , \text{v}, r = 100 , omega ) (d) ( v = 30 , \text{v}, r = 15 , \text{k}omega ) (e) ( v = 250 , \text{v}, r = 5.6 , \text{m}omega )

Explanation:

To determine the current in each case, we use Ohm's Law, which states that \( I = \frac{V}{R} \), where \( I \) is the current (in amperes, A), \( V \) is the voltage (in volts, V), and \( R \) is the resistance (in ohms, \( \Omega \)). We will calculate the current for each part (a) to (e) using this formula.

Part (a): \( V = 5 \, \text{V}, \, R = 1.0 \, \Omega \)

Step 1: Apply Ohm's Law

Using \( I = \frac{V}{R} \), substitute \( V = 5 \, \text{V} \) and \( R = 1.0 \, \Omega \).
\( I = \frac{5}{1.0} \)

Step 2: Calculate the current

\( I = 5 \, \text{A} \)

Part (b): \( V = 15 \, \text{V}, \, R = 10 \, \Omega \)

Step 1: Apply Ohm's Law

Using \( I = \frac{V}{R} \), substitute \( V = 15 \, \text{V} \) and \( R = 10 \, \Omega \).
\( I = \frac{15}{10} \)

Step 2: Calculate the current

\( I = 1.5 \, \text{A} \)

Part (c): \( V = 50 \, \text{V}, \, R = 100 \, \Omega \)

Step 1: Apply Ohm's Law

Using \( I = \frac{V}{R} \), substitute \( V = 50 \, \text{V} \) and \( R = 100 \, \Omega \).
\( I = \frac{50}{100} \)

Step 2: Calculate the current

\( I = 0.5 \, \text{A} \)

Part (d): \( V = 30 \, \text{V}, \, R = 15 \, \text{k}\Omega \)

First, convert \( 15 \, \text{k}\Omega \) to ohms: \( 15 \, \text{k}\Omega = 15 \times 10^3 \, \Omega = 15000 \, \Omega \).

Step 1: Apply Ohm's Law

Using \( I = \frac{V}{R} \), substitute \( V = 30 \, \text{V} \) and \( R = 15000 \, \Omega \).
\( I = \frac{30}{15000} \)

Step 2: Calculate the current

\( I = 2 \times 10^{-3} \, \text{A} = 2 \, \text{mA} \)

Part (e): \( V = 250 \, \text{V}, \, R = 5.6 \, \text{M}\Omega \)

First, convert \( 5.6 \, \text{M}\Omega \) to ohms: \( 5.6 \, \text{M}\Omega = 5.6 \times 10^6 \, \Omega \).

Step 1: Apply Ohm's Law

Using \( I = \frac{V}{R} \), substitute \( V = 250 \, \text{V} \) and \( R = 5.6 \times 10^6 \, \Omega \).
\( I = \frac{250}{5.6 \times 10^6} \)

Step 2: Calculate the current

\( I \approx 4.46 \times 10^{-5} \, \text{A} = 44.6 \, \mu\text{A} \)

Final Answers:

(a) \( \boldsymbol{5 \, \text{A}} \)
(b) \( \boldsymbol{1.5 \, \text{A}} \)
(c) \( \boldsymbol{0.5 \, \text{A}} \)
(d) \( \boldsymbol{2 \, \text{mA}} \) (or \( 2 \times 10^{-3} \, \text{A} \))
(e) \( \boldsymbol{\approx 44.6 \, \mu\text{A}} \) (or \( \approx 4.46 \times 10^{-5} \, \text{A} \))

Answer:

To determine the current in each case, we use Ohm's Law, which states that \( I = \frac{V}{R} \), where \( I \) is the current (in amperes, A), \( V \) is the voltage (in volts, V), and \( R \) is the resistance (in ohms, \( \Omega \)). We will calculate the current for each part (a) to (e) using this formula.

Part (a): \( V = 5 \, \text{V}, \, R = 1.0 \, \Omega \)

Step 1: Apply Ohm's Law

Using \( I = \frac{V}{R} \), substitute \( V = 5 \, \text{V} \) and \( R = 1.0 \, \Omega \).
\( I = \frac{5}{1.0} \)

Step 2: Calculate the current

\( I = 5 \, \text{A} \)

Part (b): \( V = 15 \, \text{V}, \, R = 10 \, \Omega \)

Step 1: Apply Ohm's Law

Using \( I = \frac{V}{R} \), substitute \( V = 15 \, \text{V} \) and \( R = 10 \, \Omega \).
\( I = \frac{15}{10} \)

Step 2: Calculate the current

\( I = 1.5 \, \text{A} \)

Part (c): \( V = 50 \, \text{V}, \, R = 100 \, \Omega \)

Step 1: Apply Ohm's Law

Using \( I = \frac{V}{R} \), substitute \( V = 50 \, \text{V} \) and \( R = 100 \, \Omega \).
\( I = \frac{50}{100} \)

Step 2: Calculate the current

\( I = 0.5 \, \text{A} \)

Part (d): \( V = 30 \, \text{V}, \, R = 15 \, \text{k}\Omega \)

First, convert \( 15 \, \text{k}\Omega \) to ohms: \( 15 \, \text{k}\Omega = 15 \times 10^3 \, \Omega = 15000 \, \Omega \).

Step 1: Apply Ohm's Law

Using \( I = \frac{V}{R} \), substitute \( V = 30 \, \text{V} \) and \( R = 15000 \, \Omega \).
\( I = \frac{30}{15000} \)

Step 2: Calculate the current

\( I = 2 \times 10^{-3} \, \text{A} = 2 \, \text{mA} \)

Part (e): \( V = 250 \, \text{V}, \, R = 5.6 \, \text{M}\Omega \)

First, convert \( 5.6 \, \text{M}\Omega \) to ohms: \( 5.6 \, \text{M}\Omega = 5.6 \times 10^6 \, \Omega \).

Step 1: Apply Ohm's Law

Using \( I = \frac{V}{R} \), substitute \( V = 250 \, \text{V} \) and \( R = 5.6 \times 10^6 \, \Omega \).
\( I = \frac{250}{5.6 \times 10^6} \)

Step 2: Calculate the current

\( I \approx 4.46 \times 10^{-5} \, \text{A} = 44.6 \, \mu\text{A} \)

Final Answers:

(a) \( \boldsymbol{5 \, \text{A}} \)
(b) \( \boldsymbol{1.5 \, \text{A}} \)
(c) \( \boldsymbol{0.5 \, \text{A}} \)
(d) \( \boldsymbol{2 \, \text{mA}} \) (or \( 2 \times 10^{-3} \, \text{A} \))
(e) \( \boldsymbol{\approx 44.6 \, \mu\text{A}} \) (or \( \approx 4.46 \times 10^{-5} \, \text{A} \))