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To achieve the lowest resistance when combining resistors, you should connect them in parallel. The formula for the total resistance ( R_t ) of resistors in parallel is given by ( \frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} ). For the given resistors (4 ohms, 8 ohms, 12 ohms, and 24 ohms), this results in ( \frac{1}{R_t} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24} = \frac{13}{24} ), leading to a total resistance ( R_t ) of approximately 1.85 ohms.
To find the total inductance ( L_t ) of two inductors in parallel, you can use the formula: [ \frac{1}{L_t} = \frac{1}{L_1} + \frac{1}{L_2} ] For two identical inductors of 22 mH, this simplifies to: [ \frac{1}{L_t} = \frac{1}{22 , \text{mH}} + \frac{1}{22 , \text{mH}} = \frac{2}{22 , \text{mH}} = \frac{1}{11 , \text{mH}} ] Thus, the total inductance ( L_t ) is 11 mH.
When inductors are connected in parallel, the total inductance (L_total) can be calculated using the formula: (\frac{1}{L_{total}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3}). For the given inductors, this becomes: (\frac{1}{L_{total}} = \frac{1}{0.06} + \frac{1}{0.05} + \frac{1}{0.1}). Calculating this yields (L_{total} \approx 0.017 H) or 17 mH.
When ( n ) capacitors of equal capacitance ( c ) are connected in series, the effective or equivalent capacitance ( C_{\text{eq}} ) is given by the formula: [ \frac{1}{C_{\text{eq}}} = \frac{1}{c} + \frac{1}{c} + \ldots + \frac{1}{c} = \frac{n}{c} ] Thus, the effective capacitance is: [ C_{\text{eq}} = \frac{c}{n} ] This shows that the effective capacitance decreases as the number of capacitors in series increases.
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To add the mixed numbers (3 \frac{23}{24}) and (2 \frac{3}{4}), first convert (2 \frac{3}{4}) to a fraction: (2 \frac{3}{4} = 2 + \frac{3}{4} = \frac{8}{4} + \frac{3}{4} = \frac{11}{4}). Now, convert (3 \frac{23}{24}) to an improper fraction: (3 \frac{23}{24} = \frac{72}{24} + \frac{23}{24} = \frac{95}{24}). Next, find a common denominator (which is 24) and convert (\frac{11}{4}) to (\frac{66}{24}). Finally, add the fractions: (\frac{95}{24} + \frac{66}{24} = \frac{161}{24}), which simplifies to (6 \frac{13}{24}). Thus, (3 \frac{23}{24} + 2 \frac{3}{4} = 6 \frac{13}{24}).
To add ( \frac{9}{12} ) and ( \frac{2}{4} ), first simplify ( \frac{2}{4} ) to ( \frac{1}{2} ) or ( \frac{6}{12} ) for a common denominator. Now, ( \frac{9}{12} + \frac{6}{12} = \frac{15}{12} ). This can be simplified to ( \frac{5}{4} ) or ( 1 \frac{1}{4} ).
To add ( \frac{2}{5} ) and ( \frac{2}{7} ), we first find a common denominator, which is 35. Converting the fractions, we have ( \frac{2}{5} = \frac{14}{35} ) and ( \frac{2}{7} = \frac{10}{35} ). Adding these gives ( \frac{14}{35} + \frac{10}{35} = \frac{24}{35} ). Therefore, ( \frac{2}{5} + \frac{2}{7} = \frac{24}{35} ).
To add ( \frac{8}{3} ) and ( -\frac{9}{4} ), first find a common denominator, which is 12. Rewrite the fractions: ( \frac{8}{3} = \frac{32}{12} ) and ( -\frac{9}{4} = -\frac{27}{12} ). Now add them: ( \frac{32}{12} - \frac{27}{12} = \frac{5}{12} ). Therefore, ( \frac{8}{3} + -\frac{9}{4} = \frac{5}{12} ).
To add ( \frac{3}{5} ) and ( \frac{3}{8} ), first find a common denominator, which is 40. Converting the fractions, ( \frac{3}{5} = \frac{24}{40} ) and ( \frac{3}{8} = \frac{15}{40} ). Adding these together gives ( \frac{24}{40} + \frac{15}{40} = \frac{39}{40} ). Therefore, ( \frac{3}{5} + \frac{3}{8} = \frac{39}{40} ).
To find the sum of ( \frac{3}{4} ) and ( \frac{5}{16} ), first convert ( \frac{3}{4} ) to a fraction with a denominator of 16: ( \frac{3}{4} = \frac{12}{16} ). Now, add ( \frac{12}{16} ) and ( \frac{5}{16} ): ( \frac{12}{16} + \frac{5}{16} = \frac{17}{16} ). Therefore, the sum is ( \frac{17}{16} ) or ( 1 \frac{1}{16} ).
To add ( \frac{1}{6} ) and ( \frac{3}{8} ), we first find a common denominator, which is 24. Converting the fractions, ( \frac{1}{6} ) becomes ( \frac{4}{24} ) and ( \frac{3}{8} ) becomes ( \frac{9}{24} ). Adding these gives ( \frac{4}{24} + \frac{9}{24} = \frac{13}{24} ). Thus, ( \frac{1}{6} + \frac{3}{8} = \frac{13}{24} ).
To add ( \frac{3}{4} ) and ( \frac{1}{5} ), first find a common denominator, which is 20. Convert ( \frac{3}{4} ) to ( \frac{15}{20} ) and ( \frac{1}{5} ) to ( \frac{4}{20} ). Now, add the fractions: ( \frac{15}{20} + \frac{4}{20} = \frac{19}{20} ). Thus, ( \frac{3}{4} + \frac{1}{5} = \frac{19}{20} ).
To calculate ( \frac{7}{9} - \frac{1}{4} ), you first need a common denominator, which is 36. Convert the fractions: ( \frac{7}{9} = \frac{28}{36} ) and ( \frac{1}{4} = \frac{9}{36} ). Now subtract: ( \frac{28}{36} - \frac{9}{36} = \frac{19}{36} ). Thus, ( \frac{7}{9} - \frac{1}{4} = \frac{19}{36} ).
To find ( \frac{2}{3} ) of ( \frac{3}{7} ), you multiply the two fractions: [ \frac{2}{3} \times \frac{3}{7} = \frac{2 \times 3}{3 \times 7} = \frac{6}{21}. ] Simplifying ( \frac{6}{21} ) gives ( \frac{2}{7} ). Thus, ( \frac{2}{3} ) of ( \frac{3}{7} ) is ( \frac{2}{7} ).
To subtract ( \frac{1}{6} ) from ( \frac{1}{3} ), first find a common denominator, which is 6. Convert ( \frac{1}{3} ) to ( \frac{2}{6} ). Then, subtract: ( \frac{2}{6} - \frac{1}{6} = \frac{1}{6} ). Therefore, ( \frac{1}{3} - \frac{1}{6} = \frac{1}{6} ).
To divide ( \frac{3}{5} ) by 3, you can multiply ( \frac{3}{5} ) by the reciprocal of 3, which is ( \frac{1}{3} ). This gives you ( \frac{3}{5} \times \frac{1}{3} = \frac{3}{15} ). Simplifying ( \frac{3}{15} ) results in ( \frac{1}{5} ). Therefore, ( \frac{3}{5} \div 3 = \frac{1}{5} ).