Question

1. determine the voltage drop across each resistor in figure 5 - 70.
2. three 470 ω resistors are connected in series with a 48 v source.

(a) what is the current in the circuit?
(b) what is the voltage across each resistor?
(c) what is the minimum power - rating of the resistors?

Explanation:

Step1: Calculate current for series - connected resistors

For three $470\Omega$ resistors in series, the equivalent resistance $R_{eq}=470 + 470+470=1410\Omega$. Using Ohm's law $I = \frac{V}{R}$, where $V = 48V$ and $R = R_{eq}=1410\Omega$. So, $I=\frac{48}{1410}\approx0.034A$.

Step2: Calculate voltage across each resistor

Since the resistors are in series, the current through each resistor is the same. Using Ohm's law $V = IR$ for each $470\Omega$ resistor. So, $V=0.034\times470\approx16V$.

Step3: Calculate power rating

Using the formula $P = VI$, for each resistor with $V\approx16V$ and $I = 0.034A$, $P=16\times0.034 = 0.544W$. So the minimum power - rating should be at least $0.544W$.

Answer:

(a) The current in the circuit is approximately $0.034A$.
(b) The voltage across each resistor is approximately $16V$.
(c) The minimum power rating of the resistors is at least $0.544W$.