hw - 4 begin date: 9/11/2025 12:01:00 am due date: 9/28/2025 11:00:00 pm end date: 9/28/2025 11:00:00 pm…

hw - 4 begin date: 9/11/2025 12:01:00 am due date: 9/28/2025 11:00:00 pm end date: 9/28/2025 11:00:00 pm problem 9: (6% of assignment value) if you try and measure the voltage of a battery with a voltmeter connected in series, you wont get a completely accurate measurement because of the internal resistance of the battery. to see how large this effect is, consider trying to measure the terminal voltage of a 1.585 v alkaline cell having an internal resistance of 53 ω by placing a 0.75 kω voltmeter across its terminals. randomized variables r = 0.75 kω r = 53 ω part (a) what current flows in a? grade summary deductions 0% potential 100% submissions attempts remaining: 3 (4% deduction per attempt) detailed view

hw - 4 begin date: 9/11/2025 12:01:00 am due date: 9/28/2025 11:00:00 pm end date: 9/28/2025 11:00:00 pm problem 9: (6% of assignment value) if you try and measure the voltage of a battery with a voltmeter connected in series, you wont get a completely accurate measurement because of the internal resistance of the battery. to see how large this effect is, consider trying to measure the terminal voltage of a 1.585 v alkaline cell having an internal resistance of 53 ω by placing a 0.75 kω voltmeter across its terminals. randomized variables r = 0.75 kω r = 53 ω part (a) what current flows in a? grade summary deductions 0% potential 100% submissions attempts remaining: 3 (4% deduction per attempt) detailed view

Answer

Explanation:

Step1: Recall Ohm's law formula

The formula for Ohm's law is $V = IR$, where $V$ is voltage, $I$ is current and $R$ is resistance. In the context of a battery with internal - resistance $r$, the emf $\mathcal{E}$ of the battery is related to the terminal voltage $V$, current $I$ and internal resistance $r$ by $\mathcal{E}=V + Ir$. When a voltmeter of resistance $R$ is connected across the battery terminals, the current in the circuit is given by the total resistance in the circuit. The total resistance $R_{total}=r + R$. Here, $r = 53\ \Omega$ and $R=0.75\ k\Omega=750\ \Omega$.

Step2: Calculate the current using Ohm's law

According to Ohm's law $I=\frac{\mathcal{E}}{R + r}$. Assuming the emf of the battery is equal to the terminal voltage when no current is drawn (open - circuit voltage). When the voltmeter is connected, the current $I$ in the circuit is given by $I=\frac{V}{R}$, where $V$ is the terminal voltage and $R$ is the resistance of the voltmeter. Also, from the circuit with internal resistance, if we consider the emf $\mathcal{E}$ of the battery, we know that the current $I$ in the circuit with resistance $R$ (voltmeter) and internal resistance $r$ is $I=\frac{\mathcal{E}}{R + r}$. Since we are not given the emf value and we assume the voltmeter reading gives the terminal voltage across the battery when it is connected, and we know the resistance values, we use the formula for current in a series - circuit $I=\frac{V}{R_{total}}$. Here, we assume the battery's emf is not relevant for calculating the current through the circuit formed by the voltmeter and internal resistance. The current $I$ in the circuit is calculated using $I=\frac{V}{R + r}$. Since we want to find the current flowing in the circuit with the voltmeter connected, and we know the resistance values, we use the formula $I=\frac{\mathcal{E}}{R + r}$. In the absence of information about emf, if we consider the circuit as a simple series circuit of $r$ and $R$, the current $I$ is given by $I=\frac{V}{R + r}$. Since we are not given the voltage value across the combination, we assume we are dealing with the resistance values only for the current calculation in the circuit formed by the voltmeter and internal resistance. The current $I$ in the circuit is $I=\frac{V}{R + r}$. But if we assume the battery's emf is not a factor in this part of the problem and we just consider the resistance values for the current calculation in the circuit with the voltmeter connected, we use the formula for current in a series - circuit of two resistors. The total resistance $R_{total}=r + R=53\ \Omega+750\ \Omega = 803\ \Omega$.

Step3: Calculate the current value

Using Ohm's law $I=\frac{V}{R_{total}}$, if we assume a simple case where we are just calculating the current in the circuit with the given resistances, and no other voltage information is needed for the current calculation in the circuit formed by the voltmeter and internal resistance. The current $I=\frac{V}{R + r}$. Since we are not given $V$, we can still calculate the current in terms of the resistances. The current $I$ in the circuit is $I=\frac{1.585}{750 + 53}$ (assuming the emf of the battery is $1.585\ V$ as it is the only voltage - related value given in the problem context). $I=\frac{1.585}{803}\ A\approx0.00197\ A = 1.97\ mA$.

Answer:

$I\approx1.97\ mA$