Question

An ammeter is connected in series to a battery of voltage Vb and a resistor of unknown resistance Ru(Figure 1) . The ammeter reads a current I0. Next, a resistor of unknown resistance Rr is connected in series to the ammeter, and the ammeter's reading drops to I1. Finally, a second resistor, also of resistance Rr, is connected in series as well. Now the ammeter reads I2.

Part A

If i1 / i0 = 4/5, find i2 / i0.

Express the ratio I2/I0 numerically.

An ammeter is connected in series to a battery of voltage Vb and a resistor of unknown resistance Ru(Figure 1) . The ammeter reads a current I0. Next, a resistor of unknown resistance Rr is connected in series to the ammeter, and the ammeter's reading drops to I1. Finally, a second resistor, also of resistance Rr, is connected in series as well. Now the ammeter reads I2. Part A If i1 / i0 = 4/5, find i2 / i0. Express the ratio I2/I0 numerically.

Answer 1

Concepts and reason

This problem is based on the Kirchhoff’s voltage law (KVL) and Ohm’s law.

For the calculation of the current in the circuit apply KVL in the circuit to the calculation of the current. The current in the circuit depends on the resistance of the circuit.

The current depends on the minimum resistance in the circuit.

Fundamentals

From the expression of the Kirchhoff’s voltage law (KVL) the express is as follows,

$\sum {{V_{{\rm{enclosed}}}}} = 0$

Here, ${V_{{\rm{enclosed}}}}$ is the enclosed in the circuit.

Forms Ohm’s law the expression of the voltage is expresses as follows,

$V = IR$

Here, $V$ is the voltage, $I$ is the current and $R$ is the total resistance in the circuit.

(A)

From Ohm’s law, the expression of the current ${I_0}$ in the circuit ${I_0}$ is expresses as follows,

${I_0} = \frac{{{V_{\rm{b}}}}}{{{R_{\rm{u}}}}}$

From KVL the expression of the current in the circuit ${I_1}$ is expresses as follows,

${I_1} = \frac{{{V_{\rm{b}}}}}{{{R_{\rm{u}}} + {R_{\rm{r}}}}}$

Here, ${R_{\rm{u}}} + {R_{\rm{r}}}$ is the series resistance in the circuit and ${V_{\rm{b}}}$ is the battery voltage.

The expression of the current from KVL in the circuit ${I_2}$ is expresses as follows,

${I_2} = \frac{{{V_{\rm{b}}}}}{{{R_{\rm{u}}} + 2{R_{\rm{r}}}}}$

In the above express of the current the resistance ${R_{\rm{u}}} + 2R$ is the series resistance in the circuit.

The expression for the ratio of the current $\frac{{{I_1}}}{{{I_0}}}$ is calculated as follows,

Substitute $\frac{{{V_{\rm{b}}}}}{{{R_{\rm{u}}} + {R_{\rm{r}}}}}$ for ${I_1}$ and $\frac{{{V_{\rm{b}}}}}{{{R_{\rm{u}}}}}$ for ${I_0}$ in the above expression of the current ratio,

$\begin{array}{c}\\\frac{{{I_1}}}{{{I_0}}} = \left( {\frac{{{V_{\rm{b}}}}}{{{R_{\rm{u}}} + {R_{\rm{r}}}}}} \right)\left( {\frac{1}{{\frac{{{V_{\rm{b}}}}}{{{R_{\rm{u}}}}}}}} \right)\\\\\left( {\frac{{{R_{\rm{u}}}}}{{{R_{\rm{u}}} + {R_{\rm{r}}}}}} \right) = \frac{{{I_1}}}{{{I_0}}}\\\end{array}$

Substitute $\frac{4}{5}$ for $\frac{{{I_1}}}{{{I_0}}}$ in the above equation of the current ratio,

$\begin{array}{c}\\\left( {\frac{{{R_{\rm{u}}}}}{{{R_{\rm{u}}} + {R_{\rm{r}}}}}} \right) = \frac{4}{5}\\\\5{R_{\rm{u}}} = 4\left( {{R_{\rm{u}}} + {R_{\rm{r}}}} \right)\\\\{R_{\rm{u}}} = 4{R_{\rm{r}}}\\\end{array}$

The expression of the current ratio $\frac{{{I_2}}}{{{I_0}}}$ is expresses as follows,

$\begin{array}{c}\\\frac{{{I_2}}}{{{I_0}}} = \frac{{\left( {\frac{{{V_{\rm{b}}}}}{{{R_{\rm{u}}} + 2{R_{\rm{r}}}}}} \right)}}{{\left( {\frac{{{V_{\rm{b}}}}}{{{R_{\rm{u}}}}}} \right)}}\\\\ = \frac{{{R_{\rm{u}}}}}{{{R_{\rm{u}}} + 2{R_{\rm{r}}}}}\\\end{array}$

Substitute $4{R_r}$ for ${R_{\rm{u}}}$ in the above expression of the current ratio

$\begin{array}{c}\\\frac{{{I_2}}}{{{I_0}}} = \frac{{4{R_{\rm{r}}}}}{{4{R_{\rm{r}}} + 2{R_{\rm{r}}}}}\\\\ = \frac{4}{6}\\\\ = 0.667\\\end{array}$

‎[Part A]

Part A

‎[Part A]

Ans:

The current ratio $\frac{{{I_2}}}{{{I_0}}}$ is equal to $0.667$ .