1. find the equivalent resistance and the voltage in the circuit below.\n2. find the equivalent resistance…

1.

Explanation:

Step1: Calculate equivalent resistance for series - circuit

For resistors in series, the equivalent resistance $R_{eq}$ is the sum of individual resistances. Given $R_1 = 10.0\ \Omega$, $R_2=17.0\ \Omega$, $R_3 = 4.00\ \Omega$ and $R_4=23.0\ \Omega$. $R_{eq}=R_1 + R_2+R_3+R_4=10.0 + 17.0+4.00 + 23.0=54.0\ \Omega$

Step2: Calculate voltage using Ohm's law

Ohm's law is $V = IR$. Given $I = 7.00\ A$ and $R_{eq}=54.0\ \Omega$. $V=I\times R_{eq}=7.00\times54.0 = 378\ V$

Answer:

Equivalent resistance: $54.0\ \Omega$, Voltage: $378\ V$

2.

Explanation:

Step1: Calculate equivalent resistance for parallel - circuit

The formula for resistors in parallel is $\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}$, where $R_1 = 13.0\ \Omega$, $R_2 = 7.00\ \Omega$, $R_3=15.0\ \Omega$ and $R_4 = 21.0\ \Omega$. $\frac{1}{R_{eq}}=\frac{1}{13.0}+\frac{1}{7.00}+\frac{1}{15.0}+\frac{1}{21.0}$ $\frac{1}{R_{eq}}=\frac{7\times15\times21 + 13\times15\times21+13\times7\times21 + 13\times7\times15}{13\times7\times15\times21}$ $\frac{1}{R_{eq}}=\frac{2205+4095 + 1911+1365}{28569}$ $\frac{1}{R_{eq}}=\frac{9576}{28569}$ $R_{eq}=\frac{28569}{9576}\approx2.98\ \Omega$

Step2: Calculate total current using Ohm's law

Given $V = 110\ V$ and $R_{eq}\approx2.98\ \Omega$. Using $I=\frac{V}{R_{eq}}$, we get $I=\frac{110}{2.98}\approx36.9\ A$

Answer:

Equivalent resistance: $2.98\ \Omega$, Total current: $36.9\ A$

3.

Explanation:

Step1: Use Coulomb's law formula

Coulomb's law is $F = k\frac{q_1q_2}{r^2}$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $q_1=- 28.4\times10^{-6}\ C$, $r = 16.4\times10^{-2}\ m$ and $F = 1240\ N$. We need to solve for $q_2$. Rearranging the formula gives $q_2=\frac{F\times r^{2}}{k\times q_1}$

Step2: Substitute values

$q_2=\frac{1240\times(16.4\times10^{-2})^{2}}{9\times10^{9}\times(-28.4\times10^{-6})}$ $q_2=\frac{1240\times2.6896\times10^{-2}}{9\times10^{9}\times(-28.4\times10^{-6})}$ $q_2=\frac{33.35104}{-255600}\approx - 1.30\times10^{-4}\ C$

Answer:

$q\approx - 1.30\times10^{-4}\ C$

4.

Explanation:

Step1: Use Coulomb's law for part (a)

Coulomb's law is $F = k\frac{q_1q_2}{r^2}$. The force on charge $A$ due to charge $C$, $F_{AC}=k\frac{q_Aq_C}{r_{AC}^{2}}$, and the force on charge $A$ due to charge $B$, $F_{AB}=k\frac{q_Aq_B}{r_{AB}^{2}}$. Since $q_A = q_B=q_C = 18.2\times10^{-6}\ C$, $r_{AC}=10\times10^{-2}\ m$ and $r_{AB}=(10 + 40)\times10^{-2}\ m$. $F_{AC}=9\times10^{9}\times\frac{(18.2\times10^{-6})\times(18.2\times10^{-6})}{(10\times10^{-2})^{2}}=9\times10^{9}\times\frac{331.24\times10^{-12}}{10^{-2}} = 298.116\ N$ (repulsive, to the left) $F_{AB}=9\times10^{9}\times\frac{(18.2\times10^{-6})\times(18.2\times10^{-6})}{(50\times10^{-2})^{2}}=9\times10^{9}\times\frac{331.24\times10^{-12}}{25\times10^{-2}} = 11.92464\ N$ (repulsive, to the right) The net force on $A$, $F_{net}=F_{AC}-F_{AB}=298.116 - 11.92464=286.19136\ N$ (to the left)

Step2: Use Coulomb's law for part (b)

$F_{CA}=k\frac{q_Cq_A}{r_{CA}^{2}}$, $F_{BA}=k\frac{q_Bq_A}{r_{BA}^{2}}$. $r_{CA}=10\times10^{-2}\ m$ and $r_{BA}=(10 + 40)\times10^{-2}\ m$ $F_{CA}=298.116\ N$ (repulsive, to the right) $F_{BA}=11.92464\ N$ (repulsive, to the right) The net force on $A$, $F_{net}=F_{CA}+F_{BA}=298.116+11.92464 = 310.04064\ N$ (to the right)

Answer:

(a) Net force on $A$: $286.19\ N$ (to the left) (b) Net force on $A$: $310.04\ N$ (to the right)