what is the resistance of the circuit? ω\nwhat is the voltage drop across the whole circuit? v\nthe three…

Explanation:

Step1: Calculate total resistance

Use Ohm's law $R = \frac{V}{I}$. Given $V = 1.5+1.5=3V$ and $I = 0.5A$. So $R=\frac{3}{0.5}=6\Omega$.

Step2: Find voltage - drop across whole circuit

The total voltage of the battery is $1.5V + 1.5V=3V$, so the voltage - drop across the whole circuit is $3V$.

Step3: Determine resistance of each resistor

Since the three resistors are in series and $R_{total}=R_1 + R_2+R_3$ and $R_1 = R_2 = R_3$, then $R_1=\frac{R_{total}}{3}=\frac{6}{3}=2\Omega$.

Step4: Calculate potential at position a

The positive terminal of the battery is at a higher potential. Taking the negative terminal as $0V$, the potential at position a is $3V$.

Step5: Calculate voltage - drop across first resistor

Use Ohm's law $V = IR$. For the first resistor, $I = 0.5A$ and $R = 2\Omega$, so $V=0.5\times2 = 1V$.

Step6: Calculate potential at position b

The potential at position a is $3V$, and the voltage - drop across the first resistor is $1V$. So the potential at position b is $3 - 1=2V$.

Answer:

What is the resistance of the circuit? $6\Omega$ What is the voltage drop across the whole circuit? $3V$ The three resistors are all identical. What is the resistance of each resistor? $2\Omega$ What is the potential at position a? $3V$ What is the voltage drop across the first resistor? $1V$ What is the potential at position b? $2V$